**Michal Ryszard Wojcik's PhD Thesis**

Closed and connected graphs of functions; examples of connected punctiform spaces

Katowice, Poland, 2008

MRWojcikPhD.pdf```
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\begin{document}
\title{Closed and connected graphs of functions; examples of connected
punctiform spaces}
\author{Micha{\l} Ryszard W\'{o}jcik}
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{\LARGE UNIWERSYTET \'{S}L\c{A}SKI}\\
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{\LARGE INSTYTUT MATEMATYKI}\\
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{\Large Micha{\l} Ryszard W\'{o}jcik}\\
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{\LARGE Closed and connected graphs of functions;}\\
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{\LARGE examples of connected punctiform spaces}\\
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{\Large Rozprawa doktorska}\\
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{\Large napisana pod kierunkiem}\\
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{\Large prof. dra hab. Micha{\l}a Morayne}\\
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{\LARGE Katowice 2008}
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\tableofcontents
\chapter*{Introduction}
\addcontentsline{toc}{chapter}{Introduction}
The basic intuition of a continuous function is a line drawn
from left to right without lifting one's pen off the paper.
In an attempt at capturing this intuition with mathematical formalism,
one may be tempted to think that
being continuous is the same as having a connected graph.
However, there is a discontinuous
function $\psi\colon[0,1]\to\reals$ with a connected graph:
$\psi(x)=\text{sin}(1/x)$ and $\psi(0)=0$,
which shows that the condition {\em connected graph} alone
is not subtle enough to capture our intuition.
As we look at the graph of this function we notice that it is not closed.
Indeed, it turns out that the combination {\em closed connected graph}
is a characterization of continuity for functions from the line into
locally compact spaces --- Corollary \ref{co8} in Chapter 3.
At this point there arises an intriguing question.
Does a function
$f\colon\reals^2\to\reals$ with a closed connected graph have to be continuous?
Although Ji\v{r}i Jelinek gave a negative answer in \cite{JE},
an interesting partial positive answer is the case when connectedness
of the graph follows from the assumption that all x-sections are connected
and one y-section is connected.
This is the main result of Chapter 3 given as Theorem \ref{theo7}
and published in my joint paper with M. S. W\'{o}jcik \cite{WW}.
(If $X$ is locally connected,
$Y$ is connected and locally connected, $Z$ is locally compact,
$f\colon X\times Y\to Z$ has a closed graph with all x-sections continuous
and one y-section continuous, then $f$ is continuous.)
Chapter 3 contains also an alternative version of the main result
({\em X locally connected} replaced with {\em Y locally compact}),
Theorem \ref{theonew} hitherto unpublished,
and a considerably simplified version of the original core technical lemma
--- Lemma \ref{lemma4}.
In search of a characterization of continuity that would appeal to our visual intuition of a
{\em continuously drawn line} let us notice that if we are
cutting across a strip of paper with a pair of scissors, we are in fact creating a continuous
function: the trace of the scissors is the graph.
After the whole graph is "drawn" we get two strips of paper:
the set of points above the graph and the set of points below the graph,
which leads us to the discovery that the combination of
{\em the graph is connected} and {\em the complement of the graph is disconnected}
characterizes continuity for any real-valued function defined on a connected space
--- Corollary \ref{disconnectedcomplement} in Chapter 2,
published in my joint paper with M. S. W\'{o}jcik \cite{connected_continuity}.
Coming back to the graph of our function $\psi$ let us observe that it is not locally connected.
It turns out that a function from the real line into an arbitrary space is continuous
if and only if its graph is both connected and locally connected,
which is another characterization of continuity purely in terms of connectedness of the graph
--- Theorem \ref{charka} in Chapter 2, hitherto unpublished.
Suppose a continuous real-valued function on a connected space has a local extremum
at every point without knowing whether it is a maximum or a minimum.
Does it have to be constant?
In general the answer is negative because there is a very nice, though nonmetrizable,
connected space which admits such a nonconstant function --- Example \ref{verynice}.
However, Theorem \ref{tujestkonstant} in Chapter 4 states that
in the realm of metric spaces the answer is positive for the class of
{\em separably connected spaces}, which includes all path connected spaces.
(In a separably connected space every two points can be contained
in a separable connected subset.)
It turns out that not all connected metric spaces are separably connected,
although such spaces are very hard to find.
Besides the four examples known so far,
such a space can be constructed as a dense connected graph of a function from the real line
into a nonseparable normed vector space.
In fact, we have discovered a general method for producing dense connected graphs
inside a broad class of products $X\times Y$ including any normed spaces $X,Y$
of the cardinality of the continuum --- Theorem \ref{kappa} in Chapter 4.
Although a complete connected metric space failing to be separably connected
has not been found,
a complete {\em punctiform} space can be obtained
as a closed connected graph of a function from the real line
--- Theorem \ref{Dspace}.
(A punctiform space contains no nontrivial connected compact subsets.)
In Chapter 4 we also show that for a connected subset of the unit ball
of a reflexive Banach space its being closed in the weak topology
implies being separably connected in the weak topology
--- Theorem \ref{smallbanach}.
The whole of Chapter 4 constitutes the draft of my joint paper with M. Morayne \cite{trzecia}.
In Jelinek's article certain technical details were presented in a rather cursory way
keeping some researchers in suspense whether this problem has been really solved,
therefore in Appendix A we deliver a completely rewritten construction of Jelinek's function
and a proof of its desired properties
with all the controversial details handled with painstaking care
to remove any doubts that Jelinek's original construction is indeed correct.
\chapter{Introductory Remarks}
\section{Notation and Terminology}
We define {\em extremal functions} as real-valued functions defined on topological
spaces having a local extremum at every point (without knowing whether it is a maximum
or a minimum).
A space is Baire if all of its nonempty open subsets are second category.
We define a space to be {\em strongly Baire} if all of its closed subsets are Baire.
A space is {\em totally disconnected} if singletons are its only connected subsets.
A space is {\em punctiform} if it does not contain any nontrivial compact connected subsets.
A space is {\em separably connected} if any two of its points can be contained
in a connected separable subset.
A space is {\em nonseparably connected} if it is connected and all of its nontrivial
connected subsets are nonseparable.
If $E\subset X\times Y$ then $dom(E)$ is the projection of $E$ onto $X$.
We use both $|X|$ and $card(X)$ to denote the cardinality of $X$.
When $card(X)=\kappa$, we often say that $X$ has size $\kappa$ or is of size $\kappa$.
We denote the power set of $X$ as $\mathcal{P}(X)$.
We often consider the graph of a function $F\colon X\to Y$
as a subset of the topological space $X\times Y$
or even as a topological space in its own right.
For the sake of convenience, we decided to use the capital letter $F$
rather than the usual $f$ in order to be able to treat $F$
as a subset of $X\times Y$ without having to write $Gr(f)$
or making any explanatory remarks on the spot.
When we write that $F$ is connected, we mean that $F$ is a connected
subset of $X\times Y$. Sometimes authors write $f$ is connected
to mean that $f$ maps connected sets onto connected sets
instead of the unambiguous $f$ is Darboux, but we never do that.
Let $X,Y$ be topological spaces. We say that
$f\colon X\to Y$ is a {\em Darboux} function if and only if
$f(E)$ is connected for every connected set $E\subset X$
and $f\colon X\to Y$ is a {\em connectivity} function
if and only if the graph of $f|_E$ is connected
for every connected set $E\subset X$.
\section{Separation Axioms}
Knowing that the definitions for separation axioms
are not consistently used throughout the literature,
for the sake of precision
we decided to include the following definitions,
which are equivalent with the ones given in \cite{counterexamples}.
Let $X$ be a topological space. Then
$X$ is $T_1$ if and only if all singletons are closed;
$X$ is $T_3$ if and only if for every point $x\in X$ and
every open set $U\subset X$ containing $x$ there is
an open set $V\subset X$ such that $x\in V\subset\clo{V}\subset U$;
and $X$ is $T_5$ if and only if
for any two sets $A,B\subset X$ such that
$\Clo{A}\cap B = A\cap\Clo{B} = \emptyset$,
there are two disjoint open sets
$U_1, U_2 \subset X$ such that $A\subset U_1$ and $B\subset U_2$.
Notice that being $T_3$ or $T_5$ does not imply being $T_1$.
Consider $X=\reals\times\nat$ with open sets of the form
$U\times\nat$, where $U$ is an open subset of $\reals$.
Clearly $X$ is $T_3$ and $T_5$, but $\clo{\{(x,n)\}}=\{x\}\times\nat$,
and thus no singleton is closed.
\section{Connectedness and the $T_5$ Separation Axiom}
A topological space is disconnected if and only if
it can be written as a union of two disjoint nonempty open subsets.
When we say that $E$ is a disconnected subset of
the topological space $X$ ---
by treating $E$ as a topological space with the induced topology
containing those and only those sets which are of the form $E\cap G$
where $G$ is open in $X$ ---
we obtain two open subsets $U,V$ of $X$
giving us two relatively open subsets of $E$,
namely $E\cap U$ and $E\cap V$, such that
$E=(E\cap U)\cup(E\cap V)$,
$E\cap U\not=\emptyset$ and $E\cap U\not=\emptyset$
and finally the relatively open sets are disjoint
$(E\cap U)\cap(E\cap V)=\emptyset$.
However, this does not imply that the sets $U,V$
can be chosen to be disjoint too,
which may be essential for certain arguments.
But as long as the space $X$ satisfies the $T_5$ separation axiom,
the open sets $U,V$ can be chosen to be disjoint.
Fortunately, all metric spaces are $T_5$ --- see \cite{counterexamples}.
\chapter[Continuity of Functions In Terms of Connectedness]
{Continuity of Functions In Terms of Connectedness of The Graph}
Let us take a look at the following function $\psi\colon\reals\to\reals$
with a connected graph and ask ourselves why it is not continuous.
\[\psi(x)=\begin{cases}\sin(1/x)&\ \text{if} \ x>0,\\
0&\ \text{if} \ x\leq 0.\end{cases}\]
As we look at the graph of this function,
we notice that it is not closed. It turns out that for functions $f\colon\reals\to\reals^n$
continuity is equivalent to the graph being both connected and closed.
However, the function $f(\cos(x),\sin(x))=1/x$ for $x\in(0,2\pi]$, defined on the unit circle,
is discontinuous although its graph is both closed and connected.
There are also discontinuous functions $g\colon\reals^2\to\reals$
and $h\colon\reals\to l^2$ with closed connected graphs.
A second look at the graph of our function $\psi$ reveals that it is not locally connected.
One of the results of this chapter is the observation that for functions from the real line
continuity can be characterized purely in terms of connectedness of the graph.
Namely, a function $f\colon\reals\to Y$ is continuous if and only if its graph
is both connected and locally connected.
A careful look at the function $\psi$ reveals that the complement of its graph is connected,
while the complement of the graph of any continuous real-valued function is always disconnected.
This observation leads us to the second result of this chapter, which is a characterization
of continuity for real-valued functions in terms of connectedness.
Namely, a function $f\colon X\to\reals$, defined on an arbitrary connected space $X$,
is continuous if and only if the graph is connected and its complement is disconnected.
Unfortunately, the real line cannot be reasonably replaced with a more general space
in any of the two characterizations.
\section{Connected and Locally Connected Graph}
In this section it is important to bear in mind that for functions $f\colon\reals\to Y$
having a connected graph is equivalent to being a connectivity function.
\begin{fact}
\label{connectivity}
Let $Y$ be an arbitrary topological space.
Let $F\colon\reals\to Y$ be a function with a connected graph.
Then $F$ is a connectivity function.
\end{fact}
In general, having a connected graph is powerless and almost irrelevant
when it comes to entailing connectivity.
Looking at what happens at the point of discontinuity of
the function $f(\cos(x),\sin(x))=1/x$ for $x\in(0,2\pi]$, defined on the unit circle,
serves to convince oneself
that connectedness of the graph is far from entailing the Darboux property,
not to mention connectivity.
Naturally, every connectivity function is Darboux. However, our example (at the end of this section)
of a Darboux function $f\colon(0,\infty)\to(0,\infty)$ with a totally disconnected graph shows all
too clearly that the converse is not true.
\begin{theorem}
\label{noT3}
Let $Y$ be an arbitrary topological space.
Let $F\colon\reals\to Y$ be a connectivity function.
Suppose that $F$ is locally connected at $(x,F(x))$.
Then $F$ is continuous at $x$.
\end{theorem}
\begin{proof}
Take any open set $V\subset Y$ containing $F(x)$.
Then there is a connected set $K\subset F\cap(\reals\times V)$
such that $(x,F(x))\in Int_F(K)$.
Suppose that $K\subset(-\infty,x]\times Y$.
Then the set $\{(x,F(x))\}$ is relatively clopen
in $F|_{[x,\infty)}$, which is a connected set since
$F$ is a connectivity function.
This contradiction shows that $b\in dom(K)$ for some $b>x$.
Similarly, $a\in dom(K)$ for some $a<x$.
Since $dom(K)$ is connected, the set
$U=(a,b)\subset K$ is a neighborhood of $x$ with $F(U)\subset V$,
which completes the proof that $F$ is continuous at $x$.
\end{proof}
\begin{theorem}
\label{elegancja}
Let $X$ be a locally connected space.
Let $Y$ be an arbitrary topological space.
Let $F\colon X\to Y$ be a connectivity function,
continuous at $x$.
Then $F$ is locally connected at $(x,F(x))$.
\end{theorem}
\begin{proof}
Let $E=B\times U$ be an open set containing $(x,F(x))$.
Since $F$ is continuous at $x$, there is an open set $G\subset B$
such that $x\in G$ and $F(G)\subset U$.
Since $X$ is locally connected, $G$ can be assumed to be connected.
Since $F$ is a connectivity function, the set $F|_G$ is a connected
relatively open subset of $F$ contained in $E$, which completes the proof.
\end{proof}
Putting together Fact \ref{connectivity}, Theorem \ref{noT3}, and Theorem \ref{elegancja}
we obtain a characterization of continuity purely in terms of connectedness of the graph.
\begin{theorem}
\label{charka}
Let $Y$ be an arbitrary topological space.
Let $F\colon\reals\to Y$ be a function with a connected graph.
Then $F$ is continuous at $x$ if and only if
$F$ is locally connected at $(x,F(x))$.
\end{theorem}
It would be very hard to get rid of the real line as the domain of the function
for this characterization of continuity because of the following example.
\begin{example}
There is a discontinuous function $F\colon X\to [0,1]$
with a locally arcwise connected graph
such that $F|_G$ is arcwise connected for every open connected set $G\subset X$,
and $X$ is a compact convex subset of the Euclidean plane,
thus locally connected (and as nice as can be).
\end{example}
\begin{proof}
Let $X=\{(x,y)\colon 0\leq x\leq 1\wedge 0\leq y\leq x\}$.
Let $F\colon X\to[0,1]$ be given by $F(x,0)=0$ and $F(x,y)=2xy/(x^2+y^2)$ for $y>0$.
For every $a\in[0,1]$ let $K_a=\{(x,ax)\colon x\in[0,1]\}\subset X$.
Then $F(K_a)=\{0,2a/(1+a^2)\}$.
Hence $F$ is not Darboux and F is discontinuous at $(0,0)$.
However, if $G$ is an open connected subset of $X$, then $F|_G$ is connected.
Indeed, if $G\subset X\setminus\{0,0\}$ then $F|_G$ is connected
because $F|_G$ is continuous, and if $(0,0)\in G$
then the set $F|_G=\{(0,0,0)\}\cup F|_{G\setminus\{(0,0)\}}$
is easily seen to be connected too.
It remains to show that $F$ is locally connected at $(0,0,0)$.
Take any $r>0$.
Notice that $B_r=\{(x,y)\in X\colon x<r\}$
is an open convex set containing $(0,0)$.
Then $F_r=F\cap(B_r\times[0,r))$ is an open subset of $F$ containing $(0,0,0)$.
We are done as soon as we show that $F_r$ is arcwise connected.
Consider the auxiliary function $g\colon[0,1]\to\reals$ given by $g(a)=2a/(1+a^2)$.
Notice that $g(a)<r\Longleftrightarrow a<g^{-1}(r)$.
Therefore $F^{-1}([0,r))=\bigcup\{K_a\colon 0\leq a<g^{-1}(r)\}$ is convex.
Now, the set $dom(F_r)=F^{-1}([0,r))\cap B_r$
is convex, as the intersection of two convex sets.
Finally, we conclude that $F_r$ is arcwise connected in the same way
we argue that the graph of a separately continuous function is arcwise connected.
\end{proof}
In certain arguments, functions with connected graphs defined on the real line
can be replaced with Darboux functions, which is considerably weaker
in view of the following example.
\begin{example}
There is a Darboux function $f\colon(0,\infty)\to(0,\infty)$
whose graph is totally disconnected.
\end{example}
\begin{proof}
In the first step, let us show that if $E\subset(0,\infty)$
has size $\conti$, then there is a surjection $f\colon E\to(0,\infty)$
such that $f(x)/x\not\in\ratio$ for every $x\in E$.
Observe that if $E\subset(0,\infty)$ is uncountable,
then for every $y\in(0,\infty)$ there is an $x\in E$ with $y/x\not\in\ratio$,
because otherwise $E$ would be countable.
Let us well-order the set $(0,\infty)=\{y_\alpha\colon\alpha<\conti$\}.
%,where $\conti$ is the least ordinal of the cardinality of the continuum.
We will be constructing one by one points $(x_\alpha,f(x_\alpha))$ of the graph
by transfinite induction over $\alpha<\conti$.
For $y_0$ choose some $x_0\in E$ with $y_0/x_0\not\in\ratio$ and put $f(x_0)=y_0$.
Now, for some ordinal $\beta<\conti$, the set $E\setminus\{x_\alpha\colon\alpha<\beta\}$
is of size $\conti$, so we can choose some $x_\beta\in E\setminus\{x_\alpha\colon\alpha<\beta\}$
with $y_\beta/x_\beta\not\in\ratio$ and put $f(x_\beta)=y_\beta$.
Should the set $E\setminus\{x_\alpha\colon\alpha<\conti\}$ remain nonempty after this construction
is completed, put $f(x)=\sqrt{2}$ if $x\in\ratio$ and $f(x)=1$ otherwise.
Let $\{(a_n,b_n)\colon n\in\nat\}$ be a basis for the topology of $(0,\infty)$.
Let $C_1$ be a Cantor set embedded in the open interval $(a_1,b_1)$.
Let $f\colon C_1\to(0,\infty)$ be a surjection
such that $f(x)/x\not\in\ratio$ for every $x\in C_1$.
Now, for $n+1\in\nat$, the set $(a_{n+1},b_{n +1})\setminus\bigcup_{k +1}^{n}C_k$
contains an interval $(c,d)$ because a finite union of Cantor sets is nowhere dense.
Let $C_{n+1}$ be a Cantor set embedded in $(c,d)$
and let $f\colon C_{n+1}\to(0,\infty)$ be a surjection
such that $f(x)/x\not\in\ratio$ for every $x\in C_{n+1}$.
After this construction is completed, for the remaining set
$(0,\infty)\setminus\bigcup_{n\in\nat}C_n$
put $f(x)=\sqrt{2}$ if $x\in\ratio$ and $f(x)=1$ otherwise.
Clearly, the function $f\colon(0,\infty)\to(0,\infty)$ is Darboux,
because $f([a,b])=(0,\infty)$ for any $0<a<b$. It remains to show that its graph
does not contain any connected sets with more than one point.
Let $E$ be an arbitrary subset of the graph containing two distinct points.
For some $(a,f(a))\in E$ let $\alpha=f(a)/a$.
If $E$ is contained in the line $y=\alpha x$,
it cannot be connected because then $f$ would be continuous on some interval.
Otherwise there is some $(b,f(b))\in E$ with $\beta=f(b)/b$, $\alpha\not=\beta$.
Choose a rational number $\gamma$ between $\alpha$ and $\beta$.
Now, the line $y=\gamma x$ separates $E$.
So we showed that the graph of $f$ is totally disconnected.
\end{proof}
Now that we are sensitive to the difference between a connectivity function
and a Darboux function, let us consider an alternative version of Theorem
\ref{noT3}, where the connectivity function is replaced with a Darboux function
at the cost of requiring $Y$ to be a $T_3$ space.
\begin{theorem}
\label{locally_connected_graph}
Let $Y$ be a $T_3$ space.
Let $F\colon\reals\to Y$ be a Darboux function.
If $F$ is locally connected at $(x_0,F(x_0))$,
then $F$ is continuous at $x_0$.
\end{theorem}
\begin{proof}
Take any open set $U\subset Y$ containing $F(x_0)$.
Since $F$ is locally connected at $(x_0,F(x_0))$,
there is a connected set $K\subset F$ such that
$(x_0,F(x_0))\in Int_F(K)\subset K\subset\reals\times U$.
Hence, since $Y$ is $T_3$, there is an open set $V\subset Y$
such that $F(x_0)\in V\subset\clo{V}\subset U$ and a radius $r>0$
such that $F\cap((x_0-r,x_0+r)\times\clo{V})\subset K$.
Suppose that $(x,F(x))\not\in K$ for all $x\in(x_0,x_0+r)$.
Then $F((x_0,x_0+r))\subset Y\setminus\clo{V}$.
Since $F$ is a Darboux function,
the set $A=F([x_0,x_0+r))$ is connected and covered by
the union of two disjoint open sets $V\cup(Y\setminus\clo{V})$,
each of them intersecting $A$.
This contradiction shows that there is a point $b\in(x_0,x_0+r)$
such that $b\in dom(K)$.
Since the projection $dom(K)$ is an interval, it contains $[x_0,b]$.
Similarly, there exists a point $a\in(x_0-r,x_0)$ such that
$[a,x_0]\subset dom(K)$.
So, $x_0\in (a,b)\subset dom(K)$ and $F((a,b))\subset U$,
which finishes the argument that $F$ is continuous at $x_0$.
\end{proof}
\section{Connected Graph with a Disconnected Complement}
The main theorem in this section relies on the following folklore lemma, which deserves much interest for
its own sake. It characterizes continuity by saying that $f\colon X\to\reals$ is continuous if and
only if the set of points above the graph is open and the set of points below the graph is open.
\begin{lemma}
\label{static}
Let $X$ be a topological space and $f\colon X\to\reals$.
Then $f$ is upper semicontinuous if and only if
the set $A=\{(x,y)\colon f(x)<y\}$ is open in $X\times\reals$.
Similarly, f is lower semicontinuous if and only if
the set $B=\{(x,y)\colon f(x)>y\}$ is open in $X\times\reals$.
Consequently, $f$ is continuous if and only if
the sets $A$ and $B$ are open.
\end{lemma}
The proof of this lemma is elementary.
The main theorem is given below.
\begin{theorem}
\label{main}
If $X$ is a topological space, $f\colon X\to\reals$, $Gr(f)$ is connected,
and $(X\times\reals)\setminus Gr(f)$ is disconnected then $f$ is continuous.
\end{theorem}
\begin{proof}
Since $(X\times\reals)\setminus Gr(f)$ is disconnected,
there exist two open sets $A,B\subset X\times\reals$ such that
$(X\times\reals)\setminus Gr(f)\subset A\cup B$,
$A\cap B\setminus Gr(f)=\emptyset$,
$A\setminus Gr(f)\not=\emptyset$, and $B\setminus Gr(f)\not=\emptyset$.
Notice that $A\cap B\subset Gr(f)$. Hence $A\cap B\subset Int(Gr(f))=\emptyset$.
Furthermore, since $A,B$ are disjoint open sets,
we have that $\Clo{A}\cap B=A\cap\Clo{B}=\emptyset$.
Using this and $(X\times\reals)\setminus Gr(f)\subset A\cup B$ we get that
$\Clo{A}\setminus Gr(f)\subset A$ and $\Clo{B}\setminus Gr(f)\subset B$.
Since the sets $\{x\}\times(f(x),\infty)$ and
$\{x\}\times(-\infty,f(x))$ are connected and contained in the complement of the graph
and $A,B$ are disjoint open sets whose union covers the complement of the graph,
we have the following four easy consequences:
\begin{enumerate}[(i)]
\item $(x,f(x)+1)\in A \implies \{x\}\times(f(x),\infty)\subset A$
\item $(x,f(x)+1)\in B \implies \{x\}\times(f(x),\infty)\subset B$
\item $(x,f(x)-1)\in A \implies \{x\}\times(-\infty,f(x))\subset A$
\item $(x,f(x)-1)\in B \implies \{x\}\times(-\infty,f(x))\subset B$.
\end{enumerate}
For $K,L\in\{A,B\}$, let $G^{K}_{L}=\{(x,f(x))\colon(x,f(x)+1)\in K\wedge(x,f(x)-1)\in L\}$.
Notice that since $A,B$ are disjoint, the sets $G^{A}_{ A},G^{A}_{B},G^{ B}_{A},G^{B}_{B}$
are pairwise disjoint, and since
$A$ and $B$ , we have that
$Gr(f)=G^{A}_{A}\cup G^{A}_{B}\cup G^{B}_{A}\cup G^{B}_{B}$.
We are going to show that these four sets are closed subsets of the graph,
which --- since the graph is connected --- implies that three of them are empty
and one of them is the whole graph.
Take any net $(x_t,f(x_t))$ contained in $G^{K}_{L}$ and converging to $(x,f(x))$.
Since $(x_t,f(x_t)+1)\in K$, $(x,f(x)+1)\in\Clo{K}\setminus Gr(f)\subset K$, and similarly
since $(x_t,f(x_t)-1)\in L$, $(x,f(x)-1)\in\Clo{L}\setminus Gr(f)\subset L$.
Hence $(x,f(x))\in G^{K}_{L}$, which shows that the set $G^{K}_{L}$ is closed.
Knowing that the whole graph is equal to one of these four sets
and recalling that $A\setminus Gr(f)\not=\emptyset$ and $B\setminus Gr(f)\not=\emptyset$,
we conclude that either $Gr(f)=G^{A}_{B}$ or $Gr(f)=G^{B}_{A}$.
Without loss of generality we may assume that $Gr(f)=G^{A}_{B}$.
It is now easy to notice that
\[\{(x,y)\colon f(x)<y\}\subset A,\]
\[\{(x,y)\colon f(x)>y\}\subset B.\]
Next, we are going to show that $A\cap Gr(f)=\emptyset$ and $B\cap Gr(f)=\emptyset$.
Suppose we have $(x,f(x))\in A\cap Gr(f)$. Since $A$ is open, there exist two open sets
$U\subset X$ and $V\subset\reals$ such that $(x,f(x))\in U\times V\subset A$.
Hence there exists an $\epsilon>0$ such that $(x,f(x)-\epsilon)\in A$ but
$(x,f(x)-\epsilon)\in B$ and $A\cap B=\emptyset$. This contradiction shows that
$A\cap Gr(f)=\emptyset$ and analogously $B\cap Gr(f)=\emptyset$.
Thus we have
\[\{(x,y)\colon f(x)<y\}= A,\]
\[\{(x,y)\colon f(x)>y\}= B,\]
and since $A,B$ are open, Lemma \ref{static} implies that our function is continuous.
\end{proof}
\begin{corollary}
\label{disconnectedcomplement}
Let $X$ be a connected space and $f\colon X\to\reals$.
Then $f$ is continuous if and only if $Gr(f)$ is connected
and $(X\times\reals)\setminus Gr(f)$ is disconnected.
\end{corollary}
\begin{proof}
If $Gr(f)$ is connected and $(X\times\reals)\setminus Gr(f)$ is disconnected,
then $f$ is continuous by Theorem \ref{main}. Now, if $f$ is continuous,
then $Gr(f)$ is homeomorphic with $X$ via $\Theta(x,f(x))=x$.
Hence, since $X$ is connected, $Gr(f)$ is connected. Using Lemma \ref{static}
we show that $A=\{(x,y)\colon f(x)<y\}$ and $B=\{(x,y)\colon f(x)>y\}$
are disjoint open sets covering the complement of the graph,
hence $(X\times\reals)\setminus Gr(f)$ is disconnected.
\end{proof}
It is essential that $\reals$ should be the range of the function for this characterization
of continuity because the fact that the complement of the graph of a continuous real-valued
function is disconnected follows from the following property of the real line --- it becomes
disconnected if you take away one point. It makes no sense to replace $\reals$ with a
multi-dimensional vector space because for any space $Y$ such that it is still connected after
removing any of its points the complement of the graph of any function $f\colon X\to Y$
(not necessarily continuous) is connected.
\begin{theorem}
\label{dispel}
If $X$ is a connected space, $Y$ is a topological space such that
for every $y\in Y$ the set $Y\setminus\{y\}$ is connected and
has at least two elements, and $f\colon X\to Y$ is an arbitrary function,
then $(X\times Y)\setminus Gr(f)$ is connected.
\end{theorem}
\begin{proof}
Take any open sets $A,B\subset X\times Y$ such that
$(X\times Y)\setminus Gr(f)\subset A\cup B$, $A\cap B\setminus Gr(f)=\emptyset$,
and $A\setminus Gr(f)\not=\emptyset$. Now, if we show that $B\setminus Gr(f)=\emptyset$,
we have proved that $(X\times Y)\setminus Gr(f)$ is connected.
Using the fact that the sets $\{x\}\times(Y\setminus\{f(x)\})$ are connected
we obtain the following two easy consequences:
\begin{enumerate}[(i)]
\item $(x,y)\in A\setminus Gr(f) \implies \{x\}\times(Y\setminus\{f(x)\})\subset A$
\item $(x,y)\in B\setminus Gr(f) \implies \{x\}\times(Y\setminus\{f(x)\})\subset B$.
\end{enumerate}
Let $E=\{x\in X\colon \{x\}\times(Y\setminus\{f(x)\})\subset A\}$.
Recalling that $A\setminus Gr(f)\not=\emptyset$ and keeping (i) in mind,
we conclude that $E$ is not empty. We are going to show that $E$ is closed.
Suppose that $E$ is not closed. Then there exists a point $x\in\Clo{E}\setminus E$.
Since $x\not\in E$ --- keeping (ii) in mind --- we conclude that
$\{x\}\times(Y\setminus\{f(x)\})\subset B$. There exist two distinct points
$y_1,y_2\in Y$ such that $f(x)\not=y_1$ and $f(x)\not=y_2$.
Notice that $(x,y_1)\in B$ and $(x,y_2)\in B$.
Since $B$ is open, there exist open sets $U_1,U_2\subset X$ and $V_1,V_2\subset Y$
such that $(x,y_1)\in U_1\times V_1\subset B$ and $(x,y_2)\in U_2\times V_2\subset B$.
Since $x\in U_1\cap U_2\cap\Clo{E}$, there exists a point $a\in U_1\cap U_2\cap E$.
Since $f$ is a function, either $f(a)\not=y_1$ or $f(a)\not=y_2$.
So, $f(a)\not=y_i$ for some $i\in\{1,2\}$. Since $a\in E$, $(a,y_i)\in A$.
But $(a,y_i)\in U_i\times V_i\subset B$. So $(a,y_i)$ belongs
to the empty set $A\cap B\setminus Gr(f)$. This contradiction shows that $E$ is closed.
Similarly, we show that
$F=\{x\in X\colon \{x\}\times(Y\setminus\{f(x)\})\subset B\}$ is closed.
Then $X=E\cup F$ while $E,F$ are closed and disjoint.
Since $X$ is connected and $E$ is not empty, $F$ must be empty.
This shows that $(X\times Y)\setminus Gr(f)\subset A$.
Hence $B\setminus Gr(f)=\emptyset$, which concludes the proof.
\end{proof}
\chapter{Functions with Closed Connected Graphs}
In this chapter we prove that a function $f\colon\reals\to\reals^n$ is continuous
if and only if its graph is closed and connected and
give an example of a discontinuous function $f\colon\reals\to l^2$
with a closed connected graph.
At this point it is interesting to ask whether a function
$f\colon\reals^2\to\reals$ with a closed connected graph has to be continuous.
We give a positive answer in the very special case when connectedness
of the graph follows from the assumption that all x-sections are continuous
and one y-section is continuous.
A negative answer was given by Jelinek --- see Appendix A.
\\\\
The following three theorems are known folklore results
%--- proven here in the most general setting possible ---
showing the basic properties of functions with closed graphs
into spaces with appropriate compactness properties.
\begin{theorem}
\label{theo1}
If $X,Y$ are topological spaces, $f \colon X \to Y$, $Gr(f)$ is closed,
$E \subset Y$, and $E$ is compact, then $f^{-1}(E)$ is closed.
\end{theorem}
\begin{proof}
Take any $x_0 \in \Clo{f^{-1}(E)}$. We have a net $(x_t)_{t \in \Pi}$ in
$f^{-1}(E)$ which converges to $x_0$. Notice that $(f(x_t))_{t \in \Pi}$ is
a net in the compact set $E$. Hence we get a subnet
$(f(x_{\alpha(s)}))_{s \in \Pi_0}$ which converges to some $y_0 \in E$.
Notice that $(x_{\alpha(s)},f(x_{\alpha(s)})) \to (x_0,y_0)$.
Since $Gr(f)$ is closed, $f(x_0)=y_0$.
So $f(x_0) \in E$, and $x_0 \in f^{-1}(E)$.
We showed that $\Clo{f^{-1}(E)} \subset f^{-1}(E)$,
so $f^{-1}(E)$ is closed.
\end{proof}
\begin{theorem}
\label{theo2}
If $X$ is a topological space, $Y$ is a compact space,
$f \colon X \to Y$, and $Gr(f)$ is closed, then $f$ is continuous.
\end{theorem}
\begin{proof}
Take any closed set $E \subset Y$.
Since $Y$ is a compact space, $E$ is compact.
By Theorem \ref{theo1}, $f^{-1}(E)$ is closed.
We showed that $f$ is continuous.
\end{proof}
\begin{theorem}
\label{theo3}
If $X$ is a topological space, $Y$ is a locally compact space,
$f \colon X \to Y$, and $Gr(f)$ is closed, then
$W=\{x \in X : f$ is continuous at $x \}$ is open.
\end{theorem}
\begin{proof}
Take any $x_0 \in W$. Since $Y$ is locally compact, we have
an open set $U \subset Y$ such that $f(x_0) \in U$ and
$\Clo{U}$ is compact. Since $f$ is continuous at $x_0$,
we have an open $G \subset X$ such that $x_0 \in G$ and
$f(G) \subset U$. Notice that $f|_G \colon G \to \Clo{U}$
and $Gr(f|_G)$ is closed in $G \times \Clo{U}$. By Theorem \ref{theo2},
$f|_G$ is continuous. Since $G$ is open, $f$ is continuous on $G$.
Hence $G \subset W$, and so $x_0 \in Int(W)$. We showed that
$W \subset Int(W)$, so $W$ is open.
\end{proof}
The following technical lemma is so far the most general and efficient tool
for deriving the continuity of a function with a closed graph
coupled with an appropriate connectedness condition.
\begin{lemma}
\label{lemma4}
Let $X$ be a topological space.
Let $Y$ be a locally compact space.
Let $x_0\in A\subset X$.
Suppose that $f\colon X\to Y$ has a closed graph
and $f|_A$ is continuous at $x_0$.
Suppose that for every neighborhood $G$ of $x_0$
there is a smaller neighborhood $G'$ of $x_0$ such that
for every $z\in G'$ there is a set $E\subset G$ containing $z$
such that $E\cap A\not=\emptyset$ and $f(E)$ is connected.
Then $f$ is continuous at $x_0$.
\end{lemma}
\begin{proof}
Since $Y$ is locally compact at $f(x_0)$, there is an open set $U_0\subset Y$
such that $f(x_0)\in U_0$ and $\clo{U_0}$ is compact.
Take any open set $U$ such that $f(x_0)\in U\subset U_0$.
Since $f$ has a closed graph and $\clo{U}\setminus U$ is compact,
by Theorem \ref{theo1}, there is a neighborhood $G_1$ of $x_0$
such that $f(G_1)\cap(\clo{U}\setminus U)=\emptyset$.
Since $f|_A$ is continuous at $x_0$, there is a neighborhood $G_2$ of $x_0$
such that $f(G_2\cap A)\subset U$.
Let $G=G_1\cap G_2$.
Then there is a neighborhood $G'\subset G$ of $x_0$
with the properties postulated in the last assumption.
Take any $z\in G'$.
Then $z\in E\subset G$, $a\in E\cap A$ and $f(E)$ is connected.
Since $E\subset G_1$, we can write $f(E)\subset U\cup(Y\setminus\clo{U})$,
which is a covering of a connected set by two disjoint open sets.
Since $a\in A\cap E\subset A\cap G\subset A\cap G_2$,
$f(a)\in f(E)\cap U$.
So $f(E)\subset U$ and consequently $f(z)\in U$.
Thus we showed that $f(G')\subset U$,
which completes the proof that $f$ is continuous at $x_0$.
%Now, by Theorem \ref{theo2}, $f|_{G'}\to\clo{U}$ is continuous,
%and since $G'$ is a neighborhood of $x_0$,
\end{proof}
A straightforward application of the previous highly technical lemma
yields a series of corollaries.
\begin{corollary}
\label{lemma5}
If $X$ is a topological space, $Y$ is a locally connected space,
$Z$ is a locally compact space,
$f \colon X \times Y \to Z$, $Gr(f)$ is closed, $y_0 \in Y$,
\newline
$(1) \ \text{the mapping} \ Y \ni y \mapsto f(x,y) \in Z \ \text{is Darboux for all} \ x \in X,$
\newline
$(2) \ \text{the mapping} \ X \ni x \mapsto f(x,y_0) \in Z \ \text{is continuous},$
\newline
then $f$ is continuous at $(x_0,y_0)$ for all $x_0 \in X$.
\end{corollary}
\begin{proof}
Take any $x_0 \in X$. We are preparing to apply Lemma \ref{lemma4}.
Let $A=X \times \{y_0\}$.
By (2), $f|_A$ is continuous at $(x_0,y_0)$.
Take any open sets $U$ and $V$ such that $(x_0,y_0)\in U\times V$.
Since $Y$ is locally connected, there is a connected neighborhood $K$ of $y_0$ with $K\subset V$.
Let $G'=U\times Int(K)$.
Take any $v=(x,y)\in G'$.
Let $E=\{x\}\times K$. By (1), $f(E)$ is connected.
Notice that $v\in E$ and $(x,y_0)\in E\cap A$.
Now, ready to apply Lemma \ref{lemma4}, we conclude that $f$ is continuous at $(x_0,y_0)$.
Since $x_0\in X$ was arbitrary, the proof is complete.
\end{proof}
The following corollary is a previously known result \cite{PW}.
\begin{corollary}
\label{co6}
If $X$ is a topological space, $Y$ is a locally connected space,
$Z$ is a locally compact space, $f\colon X\times Y\to Z$ has a closed graph,
\\(1) the mapping $Y\ni y\mapsto f(x,y)\in Z$ is Darboux for every $x\in X$,
\\(2) the mapping $X\ni x\mapsto f(x,y)\in Z$ is continuous for every $y\in Y$,
\\then $f$ is continuous.
\end{corollary}
By taking a singleton for $X$ in the previous corollary, we obtain the simplest
theorem about a Darboux function with a closed graph.
\begin{corollary}
\label{co7}
If $Y$ is a locally connected space, $Z$ is a locally compact space,
$f\colon Y\to Z$ is a Darboux function with a closed graph,
then $f$ is continuous.
\end{corollary}
Calling to mind the fact that a function $f\colon\reals\to Y$
with a connected graph has to be Darboux, we obtain the previously announced
characterization of continuity in terms of properties of the graph.
\begin{corollary}
\label{co8}
If $Y$ is a locally compact space and
$f\colon\reals\to Y$ has a closed connected graph
then $f$ is continuous.
\end{corollary}
The following example shows that it is essential that
$Y$ should be locally compact in Corollary \ref{co7}.
\begin{example}
\label{block}
There exists a discontinuous function $f\colon[0,1]\to l^2$
with a closed connected graph.
\end{example}
\begin{proof}
Let $Y=l^2$.
Let ${(x_{n})}_{n\in\nat}$ be a strictly decreasing sequence of positive numbers such that
\[\{x_n\colon n\in\nat\}=\{x\in(0,1]\colon\sin({\pi\over{x}})=0\}.\]
It follows that $x_1=1$. Let us define our function $f\colon[0,1]\to Y$.
Let $f(0)=(0,0,0,...)\in Y$.
Take any $x\in(0,1]=(0,x_1]$.
There is a unique $n\in\nat$ such that $x_{n+1}<x\leq x_n$.
For each $k\in\nat$ define
\[f(x)(k) =
\begin{cases}
\sin({\pi\over{x}}) \ \ \text{if} \ k=n, \\
0 \ \ \ \ \ \ \ \ \ \text{if} \ k\not=n.
\end{cases}\]
Since $f(x)$ is an element of $Y$, our function $f$ has been defined.
It is easy to see that $f|_{[x_{n+1},x_n]}$ is continuous for every $n\in\nat$.
Hence $f|_{[x_{n+1},1]}$ is continuous for every $n\in\nat$,
which means that $f$ is continuous for every $x\in(0,1]$.
Since $(x_n,f(x_n)) = (x_n,(0,0,0,...)) = (x_n,f(0)) \to (0,f(0))$
as $n\to\infty$, we conclude that $f$ has a connected graph.
Let ${(a_{n})}_{n\in\nat}$ be a strictly decreasing sequence of positive numbers such that
\[\{a_n\colon n\in\nat\}=\{x\in(0,1]\colon\sin({\pi\over{x}})=1\}.\]
Take any $n\in\nat$. Choose $k\in\nat$ so that $x_{k+1} < a_n < x_k$.
Thus $f(a_n)(k)=\sin({\pi\over{a_n}})=1$ and so $\norm{f(a_n)}=1$
for every $n\in\nat$, which means that $f$ is discontinuous at $0$.
It remains to show that $f$ has a closed graph.
Take any sequence ${(z_{n})}_{n\in\nat}$ converging to $0$ such that
${f(z_{n})}_{n\in\nat}$ converges to some point $y\in Y$.
We will show that $y=(0,0,0,...)=f(0)$. Take any $k\in\nat$.
Take any $\epsilon>0$. Since $\norm{f(z_n)-y}\to 0$ as $n\to\infty$,
there is an $n_0\in\nat$ such that
\[|f(z_n)(k)-y(k)|\leq\epsilon\] for every $n\geq n_0$.
Choose an $n\in\nat$ so that $n>n_0$ and $z_n<x_{k+1}$.
Then $f(z_n)(k)=0$. So $|y(k)|\leq\epsilon$ for every $\epsilon>0$,
and consequently $y(k)=0$ for every $k\in\nat$.
This shows that the graph of $f$ is closed.
\end{proof}
Piotrowski and Wingler noticed that a separately continuous function
$f\colon\reals\times\reals\to\reals$ with a closed graph has to be continuous --- see \cite{PW}.
The exact statement of their theorem is our Corollary \ref{co6}.
However, the assumption of separate continuity can be considerably weakened
and replaced by requiring that all x-sections are continuous
and at least one y-section is continuous.
We have even two ways of proving this in a general topological setting.
In both cases we assume that we have a function $f\colon X\times Y\to Z$
with a closed graph into a locally compact space $Z$
whose all x-sections are continuous and at least one y-sections is continuous,
with $Y$ being connected and locally connected. In the first theorem we require additionally
that $X$ be locally connected and in the second theorem we require nothing of $X$
while demanding local compactness from $Y$.
We can easily see that the condition {\em Z is locally compact} is essential by taking
$Z=\reals\times\reals$ with the river metric and $f\colon\reals\times\reals\to Z$
simply given by $f(x,y)=(x,y)\in Z$. Then all x-sections are continuous and exactly
one y-section is continuous and the graph is closed because the inverse function
is continuous. Moreover, the set of continuity points is $\reals\times\{0\}$,
which is not open, showing that local compactness is essential in Theorem \ref{theo3}.
\begin{theorem}
\label{theo7}
If
$X$ is a locally connected space,
$Y$ is a connected and locally connected space,
$Z$ is a locally compact space,
$f \colon X \times Y \to Z$, $Gr(f)$ is closed,
\newline
$(1) \
\text{the mapping} \ Y \ni y \mapsto f(x,y) \in Z \ \text{is continuous for all} \ x \in X,
$
\newline
$(2) \
\text{the mapping} \ X \ni x \mapsto f(x,y) \in Z \ \text{is continuous}
\ \text{for some} \ y \in Y
$
\newline
then $f$ is continuous.
\end{theorem}
\begin{proof}
Let $W=\{(x,y)\in X\times Y: f$ is continuous at $(x,y)\}$.
By Theorem \ref{theo3}, W is open. Take any $x_0 \in X$.
Let $D=\{y\in Y: f$ is continuous at $(x_0,y)\}$.
Notice that $D$ is open in $Y$ because $W$ is open in $X\times Y$.
We will show that $D$ is closed in $Y$.
Take any $y_0\in \Clo{D}$.
Let $\Pi=\{U \subset X \times Y: (x_0,y_0) \in U$ and $U$ is open$\}$.
Let $P=\{E \subset X \times Y : f(E)$ is connected$\}$.
Let $A=\{(x_0,y):y\in Y\}$.
Take any $U \in \Pi$. We have
open sets $G_X \subset X$, $G_Y \subset Y$ such that
$(x_0,y_0) \in G_X \times G_Y \subset U$.
Since $Y$ is locally connected, we have a connected set $K \subset G_Y$
such that $y_0 \in Int(K)$.
Since $y_0 \in \Clo{D}$, we can choose a $y' \in Int(K) \cap D$.
Since $y' \in D$, $(x_0,y') \in W$. Since $W$ is open, we have
an open set $V \subset X \times Y$ such that $(x_0,y') \in V$ and
$f$ is continuous on $V$. Now, we have open sets $V_X \subset X$,
$V_Y \subset Y$ such that $(x_0,y') \in V_X \times V_Y \subset
V \cap (G_X \times Int(K))$. Since $X$ is locally connected,
we have a connected set $T_X \subset V_X$ such that $x_0 \in Int(T_X)$.
Since $Y$ is locally connected,
we have a connected set $T_Y \subset V_Y$ such that $y' \in Int(T_Y)$.
Let $G=Int(T_X) \times Int(K)$.
$G \in \Pi$ because $x_0 \in Int(T_X)$ and $y_0 \in Int(K)$.
Take any $g=(v,z) \in G$.
Let $E=T_X \times T_Y \cup \{v\} \times K$.
We will show that $E \in P$.
Notice that $T_Y\subset K$ and $v\in T_X$.
So $T_X\times T_Y \cap \{v\}\times K \not=\emptyset$.
Hence $f(T_X\times T_Y) \cap f(\{v\}\times K)\not=\emptyset$.
Now, $T_X\times T_Y$ is connected and contained in $V$.
Since $f$ is continuous on $V$, $f(T_X\times T_Y)$ is connected.
By (1), $f(\{v\}\times K)$ is connected.
Notice that $f(E)=f(T_X\times T_Y) \cup f(\{v\}\times K)$.
Hence $f(E)$ is connected. So $E\in P$.
Notice that $(v,z) \in E$. Notice that $E\subset U$.
We have $(x_0,y')\in T_X\times T_Y \subset E$ and $(x_0,y')\in A$.
So $E\cap A\not=\emptyset$.
We showed that
\[\all_{U \in \Pi}\exi_{G \in \Pi}\all_{g \in G}\exi_{E \in P}
g \in E \wedge E \subset U \wedge E \cap A \not= \emptyset.\]
By (1), $f|_A$ is continuous at $(x_0,y_0)$.
By Lemma \ref{lemma4}, $f$ is continuous at $(x_0,y_0)$.
So $y_0\in D$. We showed that $\Clo{D}\subset D$.
So $D$ is closed in $Y$.
So $D$ is open and closed in $Y$.
By (2), we have a $y \in Y$ such that
the mapping $X \ni x \mapsto f(x,y) \in Z$ is continuous.
By Corollary \ref{lemma5}, we conclude that $f$ is continuous at $(x_0,y)$.
So $y\in D$ and $D\not=\emptyset$.
Since $Y$ is connected, $D=Y$.
Hence $f$ is continuous at $(x_0,y)$ for all $y\in Y$.
But $x_0\in X$ was arbitrary. Thus $f$ is continuous,
and the proof is complete.
\end{proof}
\begin{theorem}
\label{theonew}
If $X$ is a topological space,
$Y$ is a connected, locally connected, locally compact space,
$Z$ is a locally compact space,
$f\colon X\times Y\to Z$ has a closed graph,
\\(1) the mapping $Y\ni y\mapsto f(x_0,y)\in Z$ is Darboux for each $x_0\in X$,
\\(2) the mapping $X\ni x\mapsto f(x,y_1)\in Z$ is continuous for some $y_1\in Y$,
\\then $f$ is continuous.
\end{theorem}
\begin{proof}
By Corollary \ref{co7}, we can strengthen (1) so that
the mapping $Y\ni y\mapsto f(x_0,y)\in Z$ is continuous for each $x_0\in X$,
rather than Darboux, which will be used to map compact subsets of $Y$
onto compact subsets of $Z$.
Take any $x_0\in X$.
Let $D=\{y\in Y\colon f$ is continuous at $(x_0,y)\}$.
By Theorem \ref{theo3}, $D$ is open and by Corollary \ref{lemma5}, $y_1\in D$.
Since $Y$ is connected, we are done as soon as we show that $D$ is closed.
Let us take $y_0\in\clo{D}\setminus D$ and the proof will be finished
as soon as we reach a contradiction.
Since $Y$ is locally connected and locally compact, there is a connected
and compact neighborhood $K$ of $y_0$.
Let $E=f(\{x_0\}\times K)$. By the strengthened version of (1), $E$ is compact.
Since $Z$ is locally compact, we can obtain a closed compact set $M$
such that $E\subset Int(M)$.
Since $y_0\in\clo{D}$, there is a point $y'\in D\cap Int(K)$.
Now, $(x_0,y')$ is a point of continuity with $f(x_0,y')\in Int(M)$, so there is
a neighborhood $G$ of $x_0$ such that $f(x,y')\in Int(M)$ for every $x\in G$.
Since $(x_0,y_0)$ is a point of discontinuity, due to Theorem \ref{theo2},
there is no neighborhood $U$ of $(x_0,y_0)$ with $f(U)\subset M$.
So there is a net $(x_t,y_t)$ converging to $(x_0,y_0)$ such that
$x_t\in G$, $y_t\in Int(K)$ and $f(x_t,y_t)\in Z\setminus M$,
and consequently $f(x_t,y')\in Int(M)$.
By (1), the sets $f(\{x_t\}\times K)$ are connected.
Since each of them intersects both $Z\setminus{M}$ and $Int(M)$,
there are points $y_t'\in K$ such that $f(x_t,y_t')\in{M}\setminus Int(M)$.
Since $K$ is compact, there is a subnet $y_{\alpha(s)}'$ converging to some point $y''\in K$.
Now, the net $z_s=f(x_{\alpha(s)},y_{\alpha(s)}')$ is contained in the compact set
$M\setminus Int(M)$, and so there is another subnet $z_{\beta(w)}$
converging to some point $z_0\in M\setminus Int(M)$.
Since the graph of $f$ is closed, $f(x_0,y'')=z_0\in f(\{x_0\}\times K)=E\subset Int(M)$,
showing that $z_0\in Int(M)$, which contradicts with $z_0\in M\setminus Int(M)$.
\end{proof}
\chapter{Punctiform Spaces as Connected Graphs of Functions}
A topological space is called {\itshape separably connected}
if any two of its points can be contained in a connected separable subset.
Clearly, this is a generalization of path connectedness.
This concept arises naturally in the course of investigating the question
whether a continuous real-valued function from a connected space
having a local extremum everywhere has to be constant.
A partial positive answer is given for separably connected metric spaces.
There are easy and natural examples of nonmetrizable connected spaces
which are not separably connected.
However, connected {\itshape metric} spaces,
which are not separably connected, are hard to find.
There are four examples of such spaces in the literature
--- see \cite{POL}, \cite{SIMON}, \cite{ARON} ---
and we have constructed one more example
by using a method which turns out to have much broader applications.
In fact, we discovered a certain way of constructing functions with
connected dense graphs inside a broad class of product spaces.
Our construction is based on two ideas.
Firstly, it is inspired by Bernstein's Connected Sets
--- see \cite{counterexamples} for details.
Secondly, our space is defined as a function from the real line whose
graph is a connected dense subset of a nonseparable metric space.
Consequently, our space does not contain any nontrivial connected separable subsets
and is not locally connected at any point.
It is still an open question whether there exists a
{\em complete} connected metric space
which fails to be separably connected.
However, we obtained a {\em complete connected punctiform metric space}
as a closed connected graph of a function from the real line into a complete
metric space.
(A punctiform space contains no nontrivial compact connected subsets.)
\section{Functions with Connected Dense Graphs}
Our method for producing a connected, not separably connected metric space
is in fact a tool for constructing a function $f\colon X\to Y$
with a connected dense graph for a broad class of products $X\times Y$,
including any normed spaces $X$ and $Y$ of size $\conti$,
in which case such a function may be required to satisfy Cauchy's equation
$f(x+u)=f(x)+f(u)$. See Kulpa's paper \cite{Kulpa} for other results of this kind.
There is an old classical construction due to Leopold Vietoris of a function
$f\colon[0,1]\to[0,1]$ with a connected dense graph, \cite{Stetige_Mengen}.
\\\\
The following technical lemmas are painstakingly written so as to first of all
reveal all the details that are essential for our construction of a connected,
not separably connected metric space, and secondly get rid of all the details
which have no significance for the construction.
\begin{lemma}
\label{transfinite_construction}
Let $X,Y$ and $\mathcal{H}\subset\mathcal{P}(X\times Y)$
be arbitrary sets such that %with the cardinality constraint that
$|dom(K)|\geq|\mathcal{H}|$ for every $K\in\mathcal{H}$.
Then there exists a function $f\colon X\to Y$
which intersects every member of the family $\mathcal{H}$.
\end{lemma}
\begin{proof}
Let the ordinal $\Gamma=card(\mathcal{H})$ be used to well-order the family
$\mathcal{H}=\{K_\alpha\colon \alpha<\Gamma\}$.
We will be constructing the desired function by producing one by one
elements of its graph $(x_\alpha,f(x_\alpha))$ by transfinite induction over $\alpha<\Gamma$.
In the first step, pick some $(x_0,f(x_0))\in K_0$.
Now, given an ordinal $\beta<\Gamma$, notice that the set
$dom(K_\beta)\setminus\{x_\alpha\colon\alpha<\beta\}$
is not empty because of our cardinality constraint.
So we can safely choose some $x_\beta$ and some corresponding $f(x_\beta)$
with $(x_\beta,f(x_\beta))\in K_\beta$.
Should the set $X\setminus\{x_\alpha\colon\alpha<\Gamma\}$ remain nonempty,
fill the graph of our function in an arbitrary way just to extend the domain to the whole $X$.
\end{proof}
It is worth noting that in the following lemma $Y$ is assumed to be separably connected,
which is a nice advertisement for this little known topological property.
\begin{lemma}
\label{connected_graph}
Let $X$ be a connected space of size $\kappa$ whose each nonempty open subset
contains a closed separable set of size $\kappa$.
Let $Y$ be a separably connected $T_1$ space.
Suppose that $X\times Y$ is a $T_5$ space whose separable subsets are hereditarily separable.
Let $\mathcal{H}$ be the family of all closed separable subsets of $X\times Y$
whose projection on $X$ has size $\kappa$.
Let the function $F\colon X\to Y$ intersect every member of $\mathcal{H}$.
Then the graph of $F$ is connected and dense in $X\times Y$.
\end{lemma}
\begin{proof}
To see that $F$ is dense
notice that it intersects every set of the form $E\times\{y\}$
--- where $E$ is a closed separable set of size $\kappa$ ---
which is a set in $\mathcal{H}$ to be found in every nonempty open subset of $X\times Y$.
Suppose that $F$ is disconnected.
Since $X\times Y$ is $T_5$,
there are two disjoint open sets $U,V\subset X\times Y$ such that $F\subset U\cup V$,
$F\cap U\not=\emptyset$, $F\cap V\not=\emptyset$.
Let $A=dom(U)$ and $B=dom(V)$.
These sets, as projections of open sets, are open in $X$.
Since $F$ is a function with domain $X$, for every $x\in X$ there is a $y\in Y$
with $(x,y)\in F\subset U\cup V$, and thus $X=A\cup B$.
Since $F\cap U\not=\emptyset$, $A$ is nonempty. Similarly, $B$ is nonempty.
Since $X$ is connected it follows that $A\cap B$ is nonempty.
Hence there is a point $x_0\in X$ and two points $y_1,y_2\in Y$
with $(x_0,y_1)\in U$ and $(x_0,y_2)\in V$.
Since $U,V$ are open, there is an open set $G$ containing $x_0$
such that $G\times\{y_1\}\subset U$ and $G\times\{y_2\}\subset V$.
By assumption, there is a closed separable set $E\subset G$ of size $\kappa$.
Since $Y$ is separably connected, there is a closed connected separable set
$W\subset Y$ containing both $y_1$ and $y_2$.
Notice that the set $K=(E\times W)\setminus(U\cup V)$
is closed and separable in $X\times Y$.
To show that $E\subset dom(K)$ take any $x\in E$.
Since $(x,y_1)\in U$ and $(x,y_2)\in V$,
the connected set $\{x\}\times W$ intersects both $U$ and $V$,
which are disjoint open sets, hence it cannot be covered by their union.
Thus $(x,y)\in (\{x\}\times W)\setminus(U\cup V)\subset K$ for some $y\in Y$
ensuring that $x\in dom(K)$.
We showed that $dom(K)$ contains the set $E$ of size $\kappa$.
Thus $K\in\mathcal{H}$ and by assumption $F\cap K\not=\emptyset$.
But $F\cap K=\emptyset$.
This contradiction shows that $F$ is connected.
\end{proof}
Recall that every separable metric space can be embedded in the Hilbert cube
$[0,1]^\nat$ with the product metric, which has size $\conti$.
It is interesting to realize that there is also a similar constraint
on the cardinality of Hausdorff separable spaces, which cannot exceed $2^\conti$
--- see \cite{counterexamples} at the end of section
{\itshape Compactness Properties and the $T_i$ axioms}.
\begin{lemma}
\label{cardinality_lemma}
Let $Z$ be an arbitrary topological space.
Let $\mathcal{H}$ be the family of all closed separable subsets of $Z$.
Then $|\mathcal{H}|\leq|Z|^{\aleph_0}$.
\end{lemma}
\begin{proof}
Let $A,B$ be two distinct closed separable subsets of $Z$.
Then there are two sequences $a,b\in Z^\nat$ such that $\clo{a(\nat)}=A$ and $\clo{b(\nat)}=B$.
Since $A\not=B$, it follows that $a\not=b$.
This means that to each closed separable set we can assign a unique element of $Z^\nat$,
which completes the argument.
\end{proof}
The following theorem %, stated in a general setting,
yields the existence of functions $F\colon X\to Y$
with connected dense graphs
inside a broad class of products $X\times Y$.
\begin{theorem}
\label{kappa}
Let $X$ be a connected space whose each nonempty open subset
contains a closed separable set of size $\conti$.
Let $Y$ be a separably connected space.
Suppose that $X\times Y$ is a $T_5$ space of size $\conti$
whose separable subsets are hereditarily separable.
Then there exists a function $F\colon X\to Y$
whose graph is connected and dense in $X\times Y$.
\end{theorem}
\begin{proof}
Let $\mathcal{H}$ be the family of all closed and separable subsets of $X\times Y$
whose projection on $X$ has size $\conti$.
By Lemma \ref{cardinality_lemma},
$|\mathcal{H}|\leq\conti^{\aleph_0}=\conti$.
Thus $X,Y,\mathcal{H}$ satisfy the assumptions of
Lemma \ref{transfinite_construction}.
So there exists a function $F\colon X\to Y$
which intersects every member of $\mathcal{H}$.
By Lemma \ref{connected_graph},
the graph of $F$ is connected and dense in $X\times Y$.
\end{proof}
In the next theorem we show that if $X,Y$ are normed spaces,
$F$ may satisfy Cauchy's equation.
\begin{theorem}\label{cauchy}
Let $X$, $Y$ be normed spaces of size $\conti$.
Then there exists a function $F\colon X\to Y$
satisfying $F(x+u)=F(x)+F(u)$ for all $x,u\in X$
whose graph is connected and dense in $X\times Y$.
\end{theorem}
\begin{proof}
Let $\mathcal{H}$ be the family of all closed and separable subsets of $X\times Y$
whose projection on $X$ has size $\conti$.
By Lemma \ref{cardinality_lemma}, $|\mathcal{H}|\leq\conti^{\aleph_0}=\conti$.
If $E\subset X$ then let $\text{span}_{\ratio}(E)=
\{\Sigma_{i=1}^{n}a_i x_i\colon a_i\in\ratio,\ x_i\in E,\ i\in[1,n]\cap\nat,\ n\in\nat\}$.
Notice that $|E|<\conti\implies|\text{span}_{\ratio}(E)|<\conti$.
Let us write $\mathcal{H}=\{K_\alpha\colon \alpha<\conti\}$.
By transfinite induction over $\alpha<\conti$ we will construct a linearly independent
subset $B_0$ of vectors in $X$ over the field of rational numbers
and a function $F\colon B_0\to Y$ intersecting every member of $\mathcal{H}$.
In the step zero, we choose a nonzero vector $x_0$ such that $(x_0,F(x_0))\in K_0$.
In the $\beta$th step, we choose an $x_\beta\in dom(K_\beta)\setminus\text{span}_\ratio(\{x_\alpha\colon\alpha<\beta\})$
and some $F(x_\beta)$ such that $(x_\beta,F(x_\beta))\in K_\beta$.
Let $B_0=\{x_\alpha\colon\alpha<\conti\}$.
Let us extend $F$ to the whole $X$ in the following way.
Let $B$ be a Hamel basis for $X$ over the field of rational numbers containing the linearly independent set $B_0$.
Put $F(x)=0$ for all $x\in B\setminus B_0$ and put
$F(\Sigma_{i=1}^{n}a_i x_i)=\Sigma_{i=1}^{n}a_i F(x_i)$ for all
$a_i\in\ratio,\ x_i\in B,\ i\in[1,n]\cap\nat,\ n\in\nat$.
Now, the function $F$ satisfies $F(x+u)=F(x)+F(u)$ and by Lemma \ref{connected_graph}
its graph is connected and dense in $X\times Y$.
\end{proof}
\section{A Nonseparably Connected Metric Space}
We are now ready to show our example of a connected,
not separably connected metric space.
It will soon be clear why we had to choose
the real line for the domain of our function with a connected dense graph
to ensure that it does not contain any nontrivial connected separable subsets
and to conclude that it is not locally connected at any point.
\begin{theorem}
There exists a nonseparable connected metric space $\mathrm{M}$ of size $\conti$
with the following properties:
\begin{enumerate}[(1)]
\item each separable connected subset of $\mathrm{M}$ is a singleton,
\item $\mathrm{M}\setminus\{p\}$ is disconnected for every $p\in\mathrm{M}$,
\item $\mathrm{M}$ is not locally connected at any point.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $Y=l^{\infty}$ be the vector space of all bounded sequences of real numbers,
with the supremum norm.
Clearly $Y$ is nonseparable and of size $\conti$.
By Theorem \ref{cauchy}, we obtain a function $\mathrm{M}\colon\reals\to Y$
whose graph is connected and dense in $\reals\times Y$.
Let $E$ be a connected subset of $\mathrm{M}$ containing two distinct points,
say $a,b\in dom(E)$ with $a<b$.
Then the set $E_0=\mathrm{M}\cap([a,b]\times Y)$
is contained in $E$ because otherwise $E$ would be disconnected,
separated by $(-\infty,c)\times Y$ and $(c,\infty)\times Y$
for some $c\in(a,b)$.
Since $E_0$ is dense in $[a,b]\times Y$,
which is nonseparable, $E$ is nonseparable too.
That every point of $\mathrm{M}$ is a cut point ---
$\mathrm{M}\setminus\{p\}$ is disconnected for every $p\in\mathrm{M}$ ---
follows naturally from the fact that the domain of our function is the real line.
Being discontinuous everywhere,
$\mathrm{M}$ cannot be locally connected at any point
because of Theorem \ref{charka}.
\end{proof}
Thus we have constructed a connected metric space
whose every nondegenerate connected subset is nonseparable.
Other examples of such spaces are given in \cite{POL} and \cite{SIMON}.
\\\\
It still remains an open question whether there is a complete
connected metric space that fails to be separably connected.
We tried to prove that every complete connected metric space
has to be separably connected but instead we came up with the
following weaker result.
\begin{theorem}\label{smallbanach}
Let $X$ be a reflexive Banach space.
Suppose that $A\subset X$ is contained in the unit ball,
connected in the norm topology (which implies that it is also
connected in the weak topology) and closed in the weak topology
(which implies that it is also closed in the norm topology).
Then $A$ is separably connected in the weak topology.
\end{theorem}
\begin{proof}
Let $a,b\in A$ be distinct points.
We are going to construct a set $T\subset A$ with $a,b\in T$
that is connected and separable in the weak topology.
Take any $n\in\nat$. Since $A$ is connected, there are finitely many points
$a_0,a_1,\ldots,a_{\alpha(n)}\in A$ such that $a_0=a,a_{\alpha(n)}=b$
and $\norm{a_i-a_{i+1}}<1/n$.
Let $T_n=\bigcup_{i=1}^{\alpha (n)}\overline {a_{i-1}\ a_{i}}$,
where $\overline{u\ v}$ between $u$ and $v$.
Naturally, every $T_n$ is a compact connected set contained in the unit ball.
These sets are also compact and connected in the weak topology.
In particular, they are closed in the weak topology.
Let $\mathcal{H}$ be the collection of all subsets of the unit ball
that are closed in the weak topology.
Let $\mathcal{G}$ be the smallest topology on $\mathcal{H}$
containing sets of the form
$\{K\in\mathcal{H}\colon K\cap U\not=\emptyset\}$ and
$\{K\in\mathcal{H}\colon K\subset U\}$,
where $U\subset X$ is open in the weak topology.
The topological space $(\mathcal{H},\mathcal{G})$
is the Vietoris topology for the unit ball with respect to the weak topology.
Since $X$ is reflexive, the unit ball is compact in the weak topology
and therefore $(\mathcal{H},\mathcal{G})$ is compact --- see \cite{Michael}.
Hence we obtain a set $T\in\mathcal{H}$ which is a limit point
--- with respect to $\mathcal{G}$ --- of the sequence of sets $T_n\in\mathcal{H}$.
More precisely, $T\in\bigcap_{k\in\nat}\clo{\{T_n\colon n\ge k\}}\subset\mathcal{H}$.
Since the unit ball, being compact Hausdorff, is normal in the weak topology,
the set $T$ is connected in the weak topology --- see \cite{Michael}.
We will argue by contradiction that $a,b\in T$.
Suppose that $a\not\in T$. Let $H=X\setminus\{a\}$.
Then the set ${\bf H}=\{K\in\mathcal{H}\colon K\subset H\}$
is an open neighborhood of $T$ in $\mathcal{H}$ but
no set $T_n$ belongs to this neighborhood because they all contain $a$.
So $a\in T$ because otherwise $T$ would not be a limit point of the sequence $T_n$.
Similarly, $b\in T$.
In order to conclude that $T$ is separable
we will show that it is contained in the separable subspace
$S=\clo{\text{lin}(\bigcup_{n\in\nat}T_n)}$.
Suppose that there is a point $x\in T\setminus S$.
Since $S$ is a closed linear subspace with $x\not\in S$,
by the Hahn-Banach theorem, there is a continuous linear functional
$\Lambda\in X^{*}$ such that $\Lambda(x)=1$ and $\Lambda|_S=0$.
Let $U=\{z\in X\colon\Lambda(z)>1/2\}$.
Let ${\bf U}=\{K\in\mathcal{H}\colon K\cap U\not=\emptyset\}$.
Since $x\in T\cap U$, ${\bf U}$ is an open neighborhood of $T$ in $\mathcal{H}$.
Since $T$ is a limit point of the sequence $T_n$, there is an index $k\in\nat$
such that $T_k\in{\bf U}$ and thus $T_k\cap U\not=\emptyset$, which yields
a point $z\in S\cap U$, implying both $\Lambda(z)=0$ and $\Lambda(z)>1/2$.
This contradiction shows that $T\subset S$ and thus $T$ is separable.
It follows that $T$ is separable in the weak topology, too.
Let us argue by contradiction that $T\subset A$.
Suppose that there is a point $x\in T\setminus A$.
Now, since $A$ is closed in the weak topology,
there is a set $U$ containing $x$ and disjoint from $A$
such that $U=\bigcap_{i=1}^{N}\{z\in X\colon|\Lambda_i(z)-\Lambda_i(x)|<\epsilon\}$
where $\epsilon>0$ and $\Lambda_1,\ldots,\Lambda_N\in X^*$ with $\norm{\Lambda_i}=1$.
From the way the sets $T_n$ were constructed it follows that there is a $k\in\nat$
such that \[(\forall n\ge k)(\forall t\in T_n)(\exists a\in A)\ \norm{t-a}\le\epsilon/2.\]
Let $U'=\bigcap_{i=1}^{N}\{z\in X\colon|\Lambda_i(z)-\Lambda_i(x)|<\epsilon/2\}$
and let ${\bf V}=\{K\in\mathcal{H}\colon K\cap U'\not=\emptyset\}$.
Since $x\in T\cap U'$, ${\bf V}$ is a neighborhood of $T$ and thus there is an index
$n\ge k$ such that $T_n\in{\bf V}$, which yields a point $t\in T_n\cap U'$.
Choose a point $a\in A$ with $\norm{t-a}\le\epsilon/2$.
Since $U\cap A=\emptyset$, $a\not\in U$ and thus $|\Lambda_i(a)-\Lambda_i(x)|\ge\epsilon$ for some $i$.
Since $\norm{\Lambda_i}=1$, $|\Lambda_i(t)-\Lambda_i(a)|\le\norm{t-a}\le\epsilon/2$.
Thus, $|\Lambda_i(t)-\Lambda_i(x)|\ge|\Lambda_i(a)-\Lambda_i(x)|
-|\Lambda_i(t)-\Lambda_i(a)|\ge\epsilon-\epsilon/2=\epsilon/2$,
and consequently $t\not\in U'$.
This contradiction shows that $T\subset A$.
\end{proof}
\section{Products of Functions with Closed Connected Graphs}
We are going to construct a complete separable connected metric space with singletons
being its only connected compact subsets --- as a product of countably many functions
with closed connected graphs from the real line into a complete metric space.
For this purpose, we need to develop some tools to ensure that an appropriately chosen
sequence of functions with closed connected graphs again has a closed connected graph.
Let us first recall the notion of the product of an arbitrary family of functions
defined on the same domain. Let $X$ and $\{Y_i\}_{i\in I}$ be arbitrary sets.
Let $f_i\colon X\to Y_i$ be a family of arbitrary functions indexed with $i\in I$.
Let $Y=\Pi_{i\in I}Y_i$ be the Cartesian product of the family $\{Y_i\}_{i\in I}$.
Then the product of the family of functions $\{f_i\}_{i\in I}$ will be a function
$F\colon X\to Y$ denoted as $F=\bigotimes_{i\in I}f_i$ and defined by
%\[F(x) = \{(i,f_i(x))\colon i\in I\}\in Y\]
\[ F(x)={f_i(x)}_{i\in I} \]
for every $x\in X$.
%Obviously, the letter $F$ does not belong to the actual definition
%and was used here only for convenience of notation.
The following theorem establishes the basic facts concerning the product
of a family of functions from a topological space into topological spaces.
In particular, it serves to show that if all factor functions have closed graphs,
then their product also has a closed graph.
\begin{theorem}
\label{functionsproduct}
Let $X$ and $\{Y_i\}_{i\in I}$ be arbitrary topological spaces.
Let $f_i\colon X\to Y_i$ be arbitrary functions.
Endow $Y=\Pi_{i\in I}Y_i$ with the product topology.
Let $F\colon X\to Y$ be given by $F=\bigotimes_{i\in I}f_i$.
Then $F$ is continuous at $x_0$ if and only if
$f_i$ is continuous at $x_0$ for each $i\in I$.
Moreover, if $f_i$ has a closed graph for each $i\in I$,
then $F$ has a closed graph.
\end{theorem}
\begin{proof}
The proof is elementary so we omit it.
\end{proof}
The next two technical lemmas will be used to show that the product of
an appropriately selected family of functions with connected graphs
also has a connected graph.
\begin{lemma}
\label{pomocnik1}
Let $\{Y_i\}_{i\in I}$ be arbitrary topological spaces.
Let $f_i\colon\reals\to Y_i$ be functions with connected graphs
such that $f_i$ is discontinuous only at one point $a_i\in\reals$
%with the proviso that $a_i=a_j$ if and only if $i=j$.
with $a_i=a_j \iff i=j$.
Let $Y=\Pi_{i=1}^{\infty}Y_i$ be equipped with the product topology.
Let $F\colon\reals\to Y$ be given by $F=\bigotimes_{i\in I}f_i$.
Then $(x,F(x))\in\clo{F|_{(-\infty,x)}}\cap\clo{F|_{(x,+\infty)}}$
for every $x\in\reals$.
\end{lemma}
\begin{proof}
By Theorem \ref{functionsproduct}, $F$ is continuous on
$\reals\setminus\{a_i\colon i\in I\}$, so it suffices to write the proof
for a point of discontinuity, say $x_0=a_j$.
%Then $f_j$ is the unique function that is discontinuous at $x_0$.
Notice that $f_i$ is continuous at $x_0$ whenever $i\not=j$.
Let $g=\bigotimes_{i\in I\setminus\{j\}}f_i$.
If we write $Y^{-j}=\Pi_{i\in I\setminus\{j\}}Y_i$, then,
by Theorem \ref{functionsproduct}, the function
$g\colon X\to Y^{-j}$ is continuous at $x_0$.
Take any open set $U\subset Y$ with $F(x_0)\in U$ and any $r>0$.
We are done as soon as we find a point $a\in(x_0-r,x_0)$ with $F(a)\in U$
and a point $b\in(x_0,x_0+r)$ with $F(b)\in U$.
We may assume --- without loss of generality --- that
$U=\bigcap_{i\in I_0}\{h\in Y\colon h(i)\in U_i\}$,
where $U_i$ is open in $Y_i$ for each $i\in I_0$
and $I_0$ is a finite subset of $I$.
If $j\not\in I_0$, we may put $U_j=Y_j$ and assume that $j\in I_0$.
In any case, $f_j(x_0)\in U_j$.
Since $F(x_0)\in U$ and $g(x_0)$ is, in fact, a subset of $F(x_0)$,
it follows that $g(x_0)$ belongs to
$\bigcap_{i\in I_0\setminus\{j\}}\{h\in Y^{-j}\colon h(i)\in U_i\}$,
which is an open subset of $Y^{-j}$.
Since $g$ is continuous at $x_0$, there is a $\delta\in(0,r)$ such that
$f_i(x)\in U_i$ whenever $i\in I_0\setminus\{j\}$ and $x\in(x_0-\delta,x_0+\delta)$.
Since $f_j$ has a connected graph, there is a point $a\in(x_0-\delta,x_0)$
with $f_j(a)\in U_j$ and a point $b\in(x_0,x_0+\delta)$ with $f_j(b)\in U_j$.
Thus $F(a)\in U$ and $F(b)\in U$, which completes the proof.
\end{proof}
The following lemma is about a function from the real line whose set of all
discontinuity points has smaller cardinality than the real line.
\begin{lemma}
\label{pomocnik2}
Let $Y$ be an arbitrary topological space.
Let $F\colon\reals\to Y$ be continuous on $E$ with $|\reals\setminus E|<\conti$.
Assume that $(x,F(x))\in\clo{F|_{(-\infty,x)}}\cap\clo{F|_{(x,+\infty)}}$
for every $x\in\reals$ .
Then $F$ has a connected graph.
\end{lemma}
\begin{proof}
Suppose that $F$ is disconnected.
Then it is the union of two disjoint nonempty relatively closed sets $F_1$ and $F_2$.
Let $A=dom(F_1)$ and $B=dom(F_2)$.
Then $\reals=A\cup B$ with $A\cap B=\emptyset$ and the sets $A,B$ are nonempty.
Since $\reals$ is connected, $\clo{A}\cap\clo{B}\not=\emptyset$.
We will show that $\clo{A}\cap\clo{B}$ is contained in $\reals\setminus E$
to conclude that --- being a closed set --- it has to be countable.
Take any $x\in\clo{A}\cap\clo{B}$.
Then there are two sequences $a_n\in A$ and $b_n\in B$ converging to $x$.
If $x$ were a point of continuity, it would follow that the point $(x,F(x))$
--- being the limit of both $(a_n,F(a_n))\in F_1$ and $(b_n,F(b_n))\in F_2$ ---
belongs to $F_1\cap F_2$, which is impossible.
So the closed set $\clo{A}\cap\clo{B}$ is countable and thus has an isolated point,
say $x\in\clo{A}\cap\clo{B}$ with $(x-r,x)\cup(x,x+r)\subset
\reals \setminus (\clo{A}\cap \clo{B})=Int(\reals\setminus A)\cup Int(\reals\setminus B)
=Int(B)\cup Int(A)$.
Now, the connected set $(x,x+r)$ --- covered by the union of two disjoint open sets
$Int(A)\cup Int(B)$ --- must be contained in one of them.
The same argument goes for $(x-r,x)$.
Let us assume --- without loss of generality --- that $x\in A\cap\clo{B}$.
Then it follows that $(x-r,x)\subset B$ or $(x,x+r)\subset B$.
In either case, $(x,F(x))\in\clo{F|_{B}}$
and, as before, it follows that $(x,F(x))\in F_1\cap F_2$.
This contradiction shows that $F$ has a connected graph.
\end{proof}
\section{Completely Metrizable Connected Punctiform Space}
At the heart of our construction of a complete separable connected punctiform space
is a function from the real line into a complete metric
space with a closed connected graph and discontinuous only at one point.
That such a function exists is indeed lucky for our construction, because the condition of
having a closed connected graph is close to entailing continuity in certain natural contexts
--- see Corollary \ref{co8} in the previous chapter.
\begin{theorem}
\label{Dspace}
There exists a complete separable connected metric space $\mathrm{D}$
with the following properties:
\begin{enumerate}[(1)]
\item each connected compact subset of $\mathrm{D}$ is a singleton,
\item $\mathrm{D}\setminus\{p\}$ is disconnected for every $p\in\mathrm{D}$,
\item $\mathrm{D}$ is locally connected at each point of a dense set
$E$ with $D\setminus E$ countable.
\end{enumerate}
\end{theorem}
\begin{proof}
The basic building block of this construction is a function from the real line into some
complete metric space with a closed connected graph discontinuous at one point only.
Any such function will do.
For instance, let $f\colon\reals\to Y$ be as in Example \ref{block}, discontinuous only at 0.
Notice that the graph of $f$ is separable even if $Y$ is not.
Let $\ratio=\{a_1,a_2,\ldots\}$ with $a_i=a_j$ if and only if $i=j$.
Let $f_i\colon\reals\to Y$ be given by $f_i(x)=f(x-a_i)$ for every $i\in\nat$.
Let $Y'=Y^\nat$ with a complete product metric.
Let $F\colon\reals\to Y'$ be given by $F(x)=(f_1(x),f_2(x),\ldots)$.
By Theorem \ref{functionsproduct}, $F$ has a closed graph
and is continuous on $E=\reals\setminus\ratio$.
By Lemma \ref{pomocnik1} and Lemma \ref{pomocnik2}, $F$ has a connected graph.
We claim that the graph of $F$ with the induced metric from $\reals\times Y'$
is the desired space. Clearly, $F$ is a complete connected metric space satisfying (2).
To see that $F$ is separable let us embed it in the separable metric space
$Z=\Pi_{i=1}^{\infty}f_i$ in the following way $(x,F(x))\longleftrightarrow
\big((x,f_1(x)),(x,f_2(x)),\ldots\big)$.
Let $E$ be a connected subset of $F$ containing two distinct points
$a,b\in dom(E)$ with $a<b$. Then the set $E_0=F\cap([a,b]\times Y')$
is contained in $E$ because otherwise $E$ would be disconnected, separated
by $(-\infty,c)\times Y'$ and $(c,\infty)\times Y'$ for some $c\in(a,b)$.
Choose a rational number $q\in(a,b)$ --- a point of discontinuity of function $F$.
Then, since the graph of $F$ is closed, there is a sequence $(x_n,F(x_n))\in E_0$
such that $x_n\to q$ and $F(x_n)$ has no subsequential limit.
So $E$ is not compact.
To prove (3) we make use of our characterization of continuity for functions
from the real line with connected graphs --- Theorem \ref{charka} ---
which in our case comes down to the observation that $F$ is locally connected
at $(x,F(x))$ if and only if $F$ is continuous at $x$.
\end{proof}
The space D constructed above is a closed subset of $(l^2)^\nat$.
According to \cite{Anderson} the following three spaces are homeomorphic:
$(l^2)^\nat$, $l^2$, $\reals^\nat$.
Therefore our completely metrizable connected punctiform space
can in fact be thought of as a closed subset of $l^2$ or
as a closed subset of $\reals^\nat$.
There is an old classical example of a function $f\colon\reals\to\reals$ with
a $G_\delta$ connected punctiform graph, \cite{connexes_punctiformes}.
\section{Extremal Functions}
The concept of a separably connected space arose in the course of
investigating the question whether a continuous real-valued function
--- defined on a connected space --- having a local extremum everywhere
(without knowing for a given point whether it is a maximum or a minimum)
has to be constant.
A positive answer is given for the class of separably connected metric spaces.
However, it is rather natural for a nonmetrizable connected space to entirely
fail to be separably connected and to admit a nonconstant continuous function
with a local extremum everywhere, so the question whether such functions must
be constant has to be limited to metric spaces.
At this point it might be useful to note briefly that if $X$ is connected and
$f\colon X\to\reals$ is lower semicontinuous with a local maximum everywhere
then it must be constant because the set $\{x\in X\colon f(x)\leq f(a)\}$ is
clopen and nonempty for any $a\in X$. Therefore, in this context,
it is intriguing to study only such functions which have a local extremum
at every point without knowing whether
it is a maximum or a minimum, which will be called {\em extremal functions}.
%Notice that the function $f\colon\reals\to\reals$ given by $f(x)=1$ if $x$ is
%rational and $f(x)=0$ otherwise --- is a nonconstant extremal function.
%Also, any discontinuous step function is extremal.
%Also, the characteristic function of a finite union of arbitrary intervals
%is a nonconstant extremal function.
Naturally, in order to expect an
extremal function to be constant one has to impose some kind of connectedness
condition on its graph, but not necessarily continuity. In fact, we have shown
that every extremal Darboux function defined on a separably connected metric
space has to be constant.
It is still an open question whether a Darboux (or continuous) extremal function
defined on a connected, but not separably connected, metric space must be constant.
Any theorem in this field must not include among its assumptions both completeness
and local connectedness because of the classical theorem stating that every complete
connected and locally connected metric space is arcwise connected, and thus separably
connected --- see \cite{Engelking} section 6.3.11.
A space is {\em Baire} if all of its nonempty open subsets are of second category. Let us
call a space {\em strongly Baire} if it is Baire and all of its closed subsets are Baire.
We have proved that extremal continuous functions are constant on connected and locally
connected strongly Baire metric spaces. Although every topologically complete space
is strongly Baire, there exists a certain separable connected and locally connected
strongly Baire metric space which is not topologically complete.
\\
\\
Let us begin with a simple example which immediately brings the context
into metric spaces.
\begin{example}
\label{verynice}
There is a "very nice" nonmetrizable compact
connected linearly ordered topological space $X$,
which is not separably connected, and a continuous nonconstant function
$f\colon X\to\reals$ with a local extremum everywhere.
\end{example}
\begin{proof}
Endow $X=[0,1]\times[2,3]$ with the linear lexicographical order
and with the order topology.
The end points of $X$, namely $(0,2)$ and $(1,3)$,
cannot be contained in a separable connected subset of $X$,
because such a connected set would have to be the whole $X$,
which is nonseparable, as the open intervals
$](x,2),(x,3)[$ indexed with $x\in[0,1]$ form an uncountable
family of disjoint open nonempty sets.
Let $f\colon X\to\reals$ be given by $f(x,y)=x$.
Then $f$ is a continuous nonconstant function
with a local extremum at every point.
\end{proof}
The following theorem, combined with the fact that separable metric spaces are
second countable, constitutes the core of the argument that on a separably connected
metric space a Darboux function with a local extremum everywhere has to be constant.
\begin{theorem}
\label{fxcountable}
Let $X$ be a second countable space.
Assume that $f\colon X\to\reals$ has a local extremum everywhere.
Then $f(X)$ is countable.
\end{theorem}
\begin{proof}
For any open set $U\subset X$ let us write
\[max(U)=\{x\in U\colon f(z)\leq f(x) \text{ for all } z\in U\},\]
\[min(U)=\{x\in U\colon f(z)\geq f(x) \text{ for all } z\in U\}.\]
Notice that $f(max(U))$ and $f(min(U))$ are singletons or empty.
Let $U_1,U_2,\ldots$ be a countable basis.
Since $f$ has a local extremum everywhere,
\[X=\bigcup_{n\in\nat} max(U_n)\cup min(U_n).\]
Hence $f(X)=\bigcup_{n\in\nat}f(max(U_n))\cup f(min(U_n))$
is a countable union of singletons.
\end{proof}
Keeping the previous theorem in mind, we argue that Darboux extremal functions
are constant provided they are defined on spaces such that any two points
can be contained in a connected second countable subset.
In the case of metric spaces, we are talking about separable connectedness.
\begin{theorem}
\label{tujestkonstant}
Let $X$ be a separably connected metric space.
Let $f\colon X\to\reals$ be a Darboux function
with a local extremum everywhere.
Then $f$ is constant.
\end{theorem}
\begin{proof}
Take any two points $a,b\in X$.
Since $X$ is separably connected, there is a separable
connected set $K\subset X$ with $a,b\in K$.
Since $K$ is metrizable, it is second countable,
and by Theorem \ref{fxcountable}, $f(K)$ is countable.
Since $f$ is Darboux, $f(K)$ is connected.
As a connected countable metrizable set,
$f(K)$ is a singleton, so $f(a)=f(b)$.
Thus $f$ is constant.
\end{proof}
What follows is an attempt at producing an alternative argument
--- without recourse to separability --- that continuous extremal functions
defined on connected metric spaces ought to be constant.
\begin{theorem}
\label{denseconstant}
Let $(X,d)$ be a Baire metric space.
Let $f\colon X\to\reals$ be a continuous function with a local extremum everywhere.
Then $f$ is constant on some ball.
Moreover, $f$ is locally constant on a dense open subset.
\end{theorem}
\begin{proof}
Let\[\Delta^{max}_{r}=\{x\in X\colon(\forall z\in B(x,r))\ f(z)\leq f(x)\},\]
\[\Delta^{min}_{r}=\{x\in X\colon(\forall z\in B(x,r))\ f(z)\geq f(x)\}.\]
Notice that
\[\Delta^{max}_{r}=\bigcap_{z\in X}\{x\in X\colon d(x,z)\geq r\}
\cup\{x\in X\colon f(z)\leq f(x)\}.\]
Hence, since $f$ is continuous,
the sets $\Delta^{max}_{r}$ and $\Delta^{min}_{r}$ are closed.
Since $f$ has a local extremum everywhere,
we can write the Baire space $X$ as a countable union of closed sets
in the following way:
\[X=\bigcup_{n\in\nat}\Delta ^{max }_{1/n}\cup\bigcup_{m\in\nat}\Delta ^{min}_{1/m}.\]
So, there exist $N\in\nat$, $x_0\in X$, $r>0$ such that, say,
\makebox{$B(x_0,r)\subset \Delta^{max }_{ 1/N}$}.
Assuming that $2r<1/N$ it is easy to see that $f(B(x_0,r))\subset\{f(x_0)\}$.
Naturally, since $X$ is Baire, we can repeat this argument for any open ball,
showing that $f$ is locally constant on a dense open set.
\end{proof}
In the following theorem, we assume that the domain is strongly Baire,
which is essentially weaker than topological completeness,
even among connected and locally connected metric spaces.
\begin{theorem}
Let $X$ be a connected and locally connected metric space,
whose each closed subspace is Baire.
Let $f\colon X\to\reals$ be a continuous function with a local extremum everywhere.
Then $f$ is constant.
\end{theorem}
\begin{proof}
Let $B=\bigcup\{J\subset X\colon J$ is open and $f|_J$ is constant$\}$.
By Theorem \ref{denseconstant}, $B$ is dense.
Let us argue by contradiction that $B=X$.
Suppose that this is not the case.
Then the closed nonempty subset $A=X\setminus B$ is a Baire space.
Hence by Theorem \ref{denseconstant},
there is an open set $U\subset X$ and a point $a\in A\cap U$ such that $f(U\cap A)=\{f(a)\}$.
Since $X$ is locally connected, $U$ can be assumed to be connected.
Since $a\not\in B$, there is a point $q\in U$ such that $f(q)\not=f(a)$.
Since $f$ is continuous and since $B$ is dense, there is a point $b\in B\cap U$
such that $f(b)\not=f(a)$.
Let $B_0=\bigcup\{J\colon J$ is open and $f(J)=\{f(b)\}\}$.
Now we have the following two claims concerning the set $B_0$:
\begin{enumerate}
\item[(1)] $(\Clo{B_0}\setminus B_0)\subset A$,
\item[(2)] $(\Clo{B_0}\setminus B_0)\cap U\not=\emptyset$.
\end{enumerate}
To show that (1) holds assume that there is a point $z\in((\Clo{B_0}\setminus B_0))\setminus A$.
Then $z\in B$ and we have an open set $V$ such that $z\in V$ and $f(V)=\{f(z)\}$.
Furthermore, since $z\in\Clo{B_0}$, $f(z)=f(b)$,
and thus $V\subset B_0$ and consequently $z\in B_0$.
But $z\not\in B_0$.
To show that (2) holds assume that $(\Clo{B_0}\setminus B_0)\cap U=\emptyset$.
Then the set $U\cap B_0=U\cap\Clo{B_0}$ is clopen in $U$.
It contains $b$ but it doesn't contain $a$.
Since $a,b\in U$ and $U$ is connected, this yields a contradiction.
From (1) $\land$ (2) it follows that there is a point $z\in(A\cap U)\cap\Clo{B_0}$.
Hence $f(z)=f(a)$ and $f(z)=f(b)$, but $f(a)\not=f(b)$.
This contradiction finally shows that $X=B$.
Fix a point $b\in B$. Let $B'=$ Int$(\{x\in X\colon f(x)=f(b)\})$.
Naturally, $b\in B'$ and $B'$ is open. We will show that it is also closed.
Take any $x\in\Clo{B'}$. Since $x\in B$, there is an open set $V$ containing $x$
such that $f(V)=\{f(x)\}$.
Since $x\in\Clo{B'}$, there is a point $z\in V\cap B'$. Hence $f(z)=f(x)=f(b)$.
In effect, $f(V)=\{f(b)\}$ and thus $x\in B'$.
We showed that $\Clo{B'}\subset B'$.
So $B'$ is a nonempty clopen subset of the connected space $X$.
Hence $X=B'$ and $f$ is constant.
\end{proof}
Let us show that there exists a connected, locally connected
strongly Baire metric space that is not topologically complete.
We are going to use the fact that every separable topologically complete space
with no isolated points contains a compact set of size $\conti$.
We will find it convenient to say that a subset of a topological space is {\em sprawled}
if it intersects every compact set of size $\conti$.
Moreover, if both $A$ and $X\setminus A$ are sprawled in $X$,
then we will say that $A$ is a Bernstein subset of $X$
--- in conformance with existing terminology.
Naturally, a Bernstein subset of a separable complete metric space with no isolated points
is not topologically complete.
We will show that any subset that is sprawled in a separable complete metric space
with no isolated points is strongly Baire.
It only remains to realize that a Bernstein subset of the Euclidean plane
is connected and locally connected.
Let $X$ be a separable complete metric space with no isolated points
and let $K$ be sprawled in $X$.
To show that $K$ is strongly Baire we consider an arbitrary relatively closed subset $K_0$ of $K$
and argue that $K_0$ is Baire.
Let $K_2=\{x\in K_0\colon x \text{ is an isolated point of } K_0\}$.
Let $K_1=K_0\setminus K_2$.
$K_2$ is open in $K_0$, hence $K_1$ is closed in $K_0$.
Thus $K_1$ is closed in $K$ and has no isolated points.
Let us write $K_1=K\cap F$, where $F=\clo{K_1}=Clo_X(K_1)$.
So $F$ is closed in $X$ and has no isolated points.
Thus $F$ is separable complete with no isolated points.
Let $G_0$ be an arbitrary nonempty relatively open subset of $K_1$.
Then $G_0=K_1\cap G$, for some open subset $G$ of $X$.
Now, the set $F\cap G$ is open in $F$,
hence $F\cap G$ is topologically complete with no isolated points.
Let $A_n\subset K_1$ be an arbitrary sequence of sets which are nowhere dense in $K_1$.
Then they are nowhere dense in $F$, because $K_1\subset F$.
Actually, they are also nowhere dense in $F\cap G$.
Therefore the set $A=\bigcup_n Clo_{F\cap G}(A_n)$ has empty interior in $F\cap G$,
because $F\cap G$ is Baire.
It follows that $(F\cap G)\setminus A$ is a dense $G_\delta$ subset of $F\cap G$
with no isolated points.
Now, since $(F\cap G)\setminus A$ is topologically complete with no isolated points,
it contains a compact set of size $\conti$, say $P\subset(F\cap G)\setminus A$.
Since $K$ is sprawled in $X$, there is a point
$z\in K\cap P\subset(K\cap F\cap G)\setminus A=G_0\setminus A$,
which means that $G_0$ is not of first category in $K_1$.
Thus we showed that $K_1$ is Baire.
Since $K_2$ is a discrete space, $K_0=K_1\cup K_2$ is Baire,
completing the proof that $K$ is strongly Baire.
\appendix
\chapter[Jelinek's Discontinuous Function with Closed Connected Graph]
{Jelinek's Discontinuous Function with a Closed Connected Graph}
\section{Construction of The Graph}
\subsection*{Demarcating the Domain}
We are going to need two functions $a,r$ with the following properties:
\begin{enumerate}
\item $0\swarrow a(n)$ as $n\to\infty$
\item $a(k_1,k_2,...,k_N)\swarrow a(k_1,k_2,...,k_N,n)$ as $n\to\infty$
\item $0<r(k_1,\ldots,k_N)<min\left\{\cfrac{1}{k_1},\ldots,\cfrac{1}{k_N},\cfrac{1}{2^N}\right\}$
\item $0<a(n+1)-r(n+1)<a(n+1)+r(n+1)<\\
<a(n)-r(n)<a(n)+r(n)<1/2$
\item $a(k_1,\ldots,k_N)<\\
<a(k_1,\ldots,k_N,n +1)-r(k_1,\ldots,k_N,n+1)<\\
<a(k_1,\ldots,k_N,n +1)+r(k_1,\ldots,k_N,n+1)<\\
<a(k_1,\ldots,k_N,n )-r(k_1,\ldots,k_N,n)<\\
<a(k_1,\ldots,k_N,n )+r(k_1,\ldots,k_N,n)<\\
<a(k_1,\ldots,k_N)+ r(k_1,\ldots,k_N)/2.$
\end{enumerate}
for every $n,N\in\nat$ and every $k_1,k_2,...,k_N\in\nat$.
\\
\\
For $a=a(k_1,k_2,...,k_N)$ and $r=r(k_1,k_2,...,k_N)$ define
\[U(k_1,k_2,...,k_N)=\]
\[((a-r,r)\times(r,2^{-N}+r))\cup
( \{a\}\times(2^{-N},2^{-N}+r))\cup
((a,a+r)\times(0,2^{-N}+r)).\]
Notice that$z\in U(k_1,\ldots,k_N) \Rightarrow dist(z,\delta(U(k_1,\ldots,k_N))\leq r(k_1,\ldots,k_N)$.
\[\text{ Let }W=(0,1)^{2}\setminus \bigcup_{n=1}^{\infty}U(n).\]
\[\text{ Let }W( k_1,...,k_N)=U(k_1,...,k_N)\setminus \bigcup_{n=1}^{\infty }U(k_1,...,k_N,n).\]
Let $A^{*}=\{\ (a(k_1,...,k_N),\ y)\ \colon\ N,k_1,...,k_N\in\nat\wedge 0<y\leq 2^{-N}\ \}$.\\
Let $K=\delta([0,1]^{2})$.
Let $A=\{(k_1,...,k_N)\colon N,k_1,k_2,...,k_N\in\nat\}$. Then
\[W\cup\bigcup_{a\in A}W(a)=(0,1)^{2}.\]
\subsection*{Definition of the Function}
We are going to define a function $f\colon[0,1]^{2}\to[0,\infty)$ in the following way. Let
$f(1,y)=f(x,0)=f(x,1)=1$ for $x,y\in[0,1]$ and let
\[f(0,y)=1/y\ \ \ \text{for}\ y\in(0,1].\]
For $(a,y)\in A^{*}$ let \[f(a,y)=1/y.\]
For $(x,y)\in W\setminus A^{*}$ let
\[f(x,y)=\cfrac{1}{dist(\ (x,y),\ \delta([0,1]^{2})\ )}.\]
Notice that \[((0,1)^{2}\setminus W)\setminus A^{*}=\bigcup_{a\in A}W(a)\setminus A^{*},\]
where $\{W(a)\colon a\in A\}$ is a family of pairwise disjoint sets.
\\
\\
For $(x,y)\in W(k_1,...,k_N)$ let
\[f(x,y)=\cfrac{1}{dist(\ (x,y),\ \delta(U(k_1,...,k_N))\ )}.\]
This definition is correct because the sets $U(\ldots)$ are open.
\\
\\
We will show that this function $f\colon[0,1]^{2}\to[0,\infty)$ has a closed connected graph.
It can be extended to the whole plane by putting
\[f(x,y) =
\begin{cases}
f(-x,y), & (x,y)\in[-1,0]\times[0,1],\\
1 & (x,y)\not\in[-1,0]\times[0,1].\\
\end{cases}
\]
\section{The graph is closed}
We will split the domain of the function into several kinds of sets
and for each kind we will provide a separate proof that the graph is closed:
\[[0,1]^2=K\cup W\cup\bigcup_{a\in A}W_a=\]
\[=K\cup A^{*}\cup Int(W)\cup\bigcup_{n =1}^{\infty}\delta(U(n))
\cup\bigcup_{a\in A} [Int(W_a)\cup \bigcup_{n=1}^{\infty}\delta (U(k(a),n)]\]
\[=K\cup A^{*}\cup Int(W)\cup\bigcup_{a\in A}Int(W_a)\cup\bigcup_{a\in A}\delta(U_a).\]
\\
\\
If $(x_n,y_n)\to(x,0)$ with $y_n>0$ then $f(x_n,y_n)=y_n^{-1}$
or $f(x_n,y_n)=dist(\ldots)^{-1}$, in which case $dist(\ldots)\leq y_n$.
In any case $f(x_n,y_n)\geq y_n^{-1}$ and so $f(x_n,y_n)\to\infty$.
Hence the graph of $f$ is closed at every point $(x,0)$.
It is quite easy to see that the graph is closed at each point in
$[0,1]\times\{1\}\cup\{1\}\times[0,1]$.
\\
\\
Let $(x,y)$ be an arbitrary point in $A^{*}$: $x=a(k_1,\ldots,k_N)$ and $0<y\leq2^{-N}$.
We will examine all possible sequences $(x_n,y_n)$ converging to $(x,y)$:
\begin{enumerate}
\item if $(x_n,y_n)\in A^{*}$ then $f(x_n,y_n)={y_n}^{-1}\to y^{-1}=f(x,y)$,
\item if $(x_n,y_n)\in W(k_1,\ldots,k_N)\setminus A^{*}$ then\\
$f(x_n,y_n)={dist(\ (x,y),\ \delta(U(k_1,...,k_N))\ )}^{-1}={|x_n-x|}^{-1}\to\infty$,
\item if $(x_n,y_n)\in W(k_1,\ldots,k_N,k_{N+1}(n),\ldots)$ then\\
$f(x_n,y_n)={dist(\ (x_n,y_n),\ \delta(U(k_1,...,k_N,k_{N+1}(n),\ldots))\ )}^{-1}\\
\geq{r(k_1,\ldots,k_N,k_{N+1}(n)},\ldots)^{-1}\geq k_{N+1}(n)\to\infty$,
\item if $(x_n,y_n)\in W$ or $(x_n,y_n)\in W(k_1,\ldots,k_{N-1})$
then $f(x_n,y_n)\to f(x,y)$.
\end{enumerate}
Hence the graph of $f$ is closed at every point in $A^{*}$.
Analogously, we convince ourselves that the graph is closed at every point $(0,y)$.
Thus the graph is closed at every point in $K\cup A^{*}$.
Furthermore, if $z\in Int(W)\cup\bigcup_{a\in A}Int(W_a)$
then $f$ is clearly continuous at $z$, hence closed.
\\
\\
Let $z\in\delta U_a\setminus A^{*}$ for some $a\in A$.
We will examine all possible sequences $z_n$ converging to $z$.
If $z_n\in U_a$ then $f(z_n)={dist(z_n,\delta U_a)}^{-1}\to\infty$,
and if $z_n\not\in U_a$ then $f(z_n)\to f(z)$.
Hence the graph is closed at $z$.
Thus we have completed the proof that the graph of $f$ is closed.
\newpage
\section{The graph is connected}
The proof that the graph of $f$ is connected is based on the following scheme
\begin{enumerate}
\item $[0,1]^{2}=K\cup W\cup\bigcup_{a\in A}W_a$
\item $Gr(f|_K)\cup Gr(f|_W)$ is connected
\item $Gr(f|_W)\cup Gr(f|_{W(n)})$ is connected
\item $Gr(f|_{W(k_1,\ldots,k_N)})\cup Gr(f|_{W(k_1,\ldots,k_N,n)})$ is connected
\end{enumerate}
for every $N\in\nat$, every $k_1,\ldots,k_N\in\nat$, and every $n\in\nat$.
\\
\\
Evidently, $Gr(f|_K)$ is connected. We will show that the sets
$Gr(f|_W)$, $Gr(f|_{W(k_1,\ldots,k_N)})$ are connected
by pointing out that the sets $W$, $W(k_1,\ldots,k_N)$ are connected
and proving that $f$ restricted to any of these sets is continuous.
Points (2)-(4) will be proved by using the following lemma.
\begin{lemma}
If $A,B$ are connected and $A\cap\Clo{B}\not=\emptyset$ then $A\cup B$ is connected.
\end{lemma}
Notice that
\[Gr(f|_K)\ni(0,1/2,2)=lim_{n\to\infty}(a(n),1/2,2),\]
\[(a(n),1/2,2)\in Gr(f|_W).\]
Hence \[(0,1/2,2)\in Gr(f|_K)\cap\Clo{Gr(f|_W)}.\]
So $Gr(f|_K)\cup Gr(f|_W)$ is connected.\\
Similarly, we show that
$Gr(f|_W)\cup Gr(f|_{W(k_1)})$ is connected for each $k_1\in\nat$:
\[Gr(f|_W)\ni(a(k_1),1/4,4)=lim_{n\to\infty}(a(k_1,n),1/4,4),\]
\[(a(k_1,n),1/4,4)\in Gr(f|_{W(k_1)}).\]
Hence \[(a(k_1),1/4,4)\in Gr(f|_W)\cap\Clo{Gr(f|_{W(k_1)})}.\]
So $Gr(f|_W)\cup{Gr(f|_{W(k_1)})}$ is connected.
Thus points (2) and (3) have been demonstrated and (4) can be done analogously.
To complete the proof that the graph of $f$ is connected we still need to
prove the continuity of $f|_W$ and $f|_{W(k_1,\ldots,k_N)}$.
We will skip $f|_W$ and focus on $f|_{W(k_1,\ldots,k_N)}$
because $f|_W$ can be done analogously.
\\
\\
We are going to show that $f$ restricted to ${W(k_1,\ldots,k_N)}$
is continuous by using the following lemma.
\begin{lemma}
If $E,H$ are closed, $g\colon E\cup H\to Y$ and $g|_{E}$, $g|_{H}$ are continuous
then $g$ is continuous.
\end{lemma}
Let us define the two sets $H$ and $E$ as follows:
\[H=\bigcup_{n=1}^{\infty}\{a(k_1,\ldots,k_N,n)\}\times
[a(k_1,\ldots ,k_N,n)-a(k_1,\ldots ,k_N),2^{-N}],\]
\[E=W(k_1,\ldots,k_N)\setminus \bigcup_{n=1}^{\infty }H_n\text{, where}\]
\[H_n=\{a(k_1,\ldots ,k_N,n)\}\times (a(k_1,\ldots ,k_N,n) -a(k_1,\ldots,k_N),2^{-N}].\]
Notice that $H$ is a relatively closed subset of $W(k_1,\ldots,k_N)$.
We will show that each $H_n$ is a relatively open subset of $W(k_1,\ldots,k_N)$
in order to conclude that $E$ is a relatively closed subset of $W(k_1,\ldots,k_N)$.
Obviously, $W(k_1,\ldots,k_N)=E\cup H$.
Notice that $f|_H(x,y)=y^{-1}$, hence $f|_{H}$ is continuous.
We will show that $f|_E(x,y)=dist(\ (x,y),\ \delta(U(k_1,...,k_N))\ )^{-1}$
to conclude that $f|_E$ is continuous.
All of this put together implies that $f$ restricted to $W(k_1,\ldots,k_N)$
is continuous.
The reasoning above is made complete by the following claims and their proofs.
\begin{claim} The set
\[H_n=\{a(k_1,\ldots,k_N,n)\}\times (a(k_1,\ldots,k_N,n)-a(k_1,\ldots,k_N),2^{-N}]\]
is a relatively open subset of $W(k_1,\ldots,k_N)$.
\end{claim}
\begin{proof}
Let us simplify notation as follows. Let $a'=a(k_1,\ldots,k_N)$.\\
Let $a=a(k_1,\ldots,k_N,n)$. Let $r=r(k_1,\ldots,k_N,n)$.\\
Let $W_a=W(k_1,\ldots,k_N)$. Let $U_a=U(k_1,\ldots,k_N,n)$.\\
Let $(a,y_0)$ be an arbitrary element of $H_n$.
Choose $y'$ so that $a-a'<y'<y_0$.
Let $G=(a-r,a+r)\times (y',y_0+r)$.
Our proof will be finished as soon as we show that $G\cap W_a\subset H_n$.
Take any $(x,y)\in G\cap W_a$. Now, $a-r<x<a+r$.
Since $(x,y)\in W_a$ it follows that $(x,y)\not\in U_a$.
Notice that
\[r<a-a'<y'<y<y_0+r\leq 2^{-N}+r.\]
Now, if $a-r<x<a$ then
$(x,y)\in(a-r,a)\times(r,2^{-N}+r)\subset U_a$.
On the other hand, if $a<x<a+r$ then
$(x,y)\in(a,a+r)\times(0,2^{-N}+r)\subset U_a$.
Thus $x=a$. If $y>2^{-N}$ then $(x,y)\in\{a\}\times(2^{-N},2^{-N}+r)\subset U_a$.
Thus $y\leq 2^{-N}$. So $(x,y)\in H_n$.
\end{proof}
The following proposition will be used to show that\\
$f|_E(x,y)=dist(\ (x,y),\ \delta(U(k_1,...,k_N))\ )^{-1}$.
\begin{claim}
If $0<y<a(k_1,\ldots,k_N,n)-a(k_1,\ldots,k_N)$ then
\[y=dist(\ (a(k_1,\ldots,k_N,n),y),\ \delta(U(k_1,\ldots,k_N))\ ).\]
\end{claim}
\begin{proof}
Let us simplify notation as follows.\\
Let $a'=a(k_1,\ldots,k_N)$. Let $r'=r(k_1,\ldots,k_N)$.\\
Let $a=a(k_1,\ldots,k_N,n)$. Let $r=r(k_1,\ldots,k_N,n)$.\\
Let $U_{a'}=U(k_1,\ldots,k_N)$.\\
Recall that $a-a'<r'<2^{-N}$.
Hence \[0<y<\cfrac{r'+2^{-N}}{2}\ \ \text{ and }\ \ 0<y<2^{-N}.\]
Recall that $a'<a<a'+r'/2$.
Considering the shape of the boundary of $U_{a'}$,
these three inequalities allow us to conclude that
\[dist((a,y),\delta(U_{a'}))=min\{a-a',y\}=y.\]
\end{proof}
\begin{claim}
$f|_E(x,y)=\cfrac{1}{dist(\ (x,y),\ \delta(U(k_1,\ldots,k_N))\ )}$.
\end{claim}
\begin{proof}
Recall that $E\subset W(k_1,\ldots,k_N))$.
Hence, for $(x,y)\in E\setminus A^{*}$
--- by the definition of $f$ --- we have
$f|_E(x,y)={dist(\ (x,y),\ \delta(U(k_1,\ldots,k_N))\ )}^{-1}$.
Notice that
\[E\cap A^{*}=\bigcup_{n=1}^{\infty}\{a(k_1,\ldots,k_N,n)\}\times
(0,a(k_1,\ldots,k_N,n)-a(k_1,\ldots,k_N)].\]
We finish this proof by referring to the previous claim.\end{proof}
\section{The graph is not arcwise connected}
Let $L=f|_{\{0\}\times(0,1/2]}$.
Notice that $f|_{K}\setminus L$ is open in $Gr(f)$.
Hence \[(1)\ f|_K\cap\Clo{f\setminus f|_K}\subset L.\]
Let us choose some $z_0\in f|_K$ and $z_1\in f\setminus f|_K$.
If the graph of $f$ were arcwise connected then there would exist
a continuous function $h\colon[0,1]\to Gr(f)$ such that
$h(0)=z_0$ and $h(1)=z_1$. We will suppose that such a function
exists and obtain a contradiction, thus proving that the graph
is not arcwise connected. Let $E=h([0,1])$.
The set $E$ has the following properties:
\[(2)\ E \text{ is bounded and connected,}\]
\[(3)\ E\cap f|_K\not=\emptyset \text{ and } E\setminus f|_K\not=\emptyset.\]
Moreover, we will show that
\[(4)\ E\cap f|_K\cap\Clo{E\setminus f|_K}\not=\emptyset.\]
Let us argue by contradiction.
If $E\cap f|_K\cap\Clo{E\setminus f|_K}=\emptyset$, then
$E\cap f|_K\subset E\setminus\Clo{E\setminus f|_K}\subset E\cap f|_K$.
Hence $E\cap f|_K=E\setminus\Clo{E\setminus f|_K}$ is a
relatively closed and relatively open subset of the connected set $E$.
By (3), it is not empty and it is a proper subset of $E$.
To obtain this contradiction
we have used the fact that $f|_K$ is a relatively closed subset of $Gr(f)$.
Now, putting (1) and (4) together we obtain a point $z$ such that
\[(5)\ z\in E\cap f|_K\cap\Clo{E\setminus f|_K}\cap L.\]
Independently of the reasoning above,
there exists a $\delta>0$ such that
\[(6)\ E\cap f|_{(0,\delta)\times(0,1)}\subset f|_{A^{*}}\]
because otherwise the set $E$ would not be bounded.
From (5) and (6) it follows that
there exists a point $(a,y)\in A^{*}\cap(0,\delta)\times(0,1)$
with $((a,y),f(a,y))\in E$.
Furthermore, there exists an $r>0$ such that $a+r<\delta$ and
\[(7)\ A^{*}\cap[a,a+r]\times (0,1)=A^{*}\cap(a-r,a+r)\times(0,1).\]
By (6) and (7),
\[E\cap f|_{[a,a+r]\times(0,1/2]}=E\cap f|_{(a-r,a+r)\times(0,1)}.\]
Notice that the set above is nonempty because it contains $((a,y),f(a,y))$.
Moreover, it is a relatively closed and relatively open
proper subset of the connected set $E$. This contradiction completes the proof.
%Quite similarly, this reasoning can be used to prove that the graph is not
%locally connected at any point in $L$.
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