Math ASCII Notation Demo

Mathematical content on Apronus.com is presented in Math ASCII Notation which can be properly displayed by all Web browsers because it uses only the basic set of characters found on all keyboards and in all fonts.

The purpose of these pages is to demonstrate the power of the Math ASCII Notation. In principle, it can be used to write mathematical content of any complexity. In practice, its limits can be seen when trying to write complicated formulas (containing, for example, variables with many indexes or multiple integrals).

Despite its limitations the Math ASCII Notation has much expressive power, as can be seen from browsing through these pages.

Let K be the set of all functions f:|N->|R such that /\n:-|N 0<=f(n)<=1/n.
Show that if a sequence of functions from K converges pointwise, then it converges uniformly.
page 157 in palace
Prove that every sequence of functions f:|N->[0,1] has a subsequence that converges pointwise.
page 158 in palace
Let d be a metric on P(N) defined by d(A,B)=1/min(A+B).
Let E = { f:N->R : \/AcN /\x:-N f(x)=1/x * 1A(x) }.
Show that (P(N),d) and (E,sup) are isometric spaces.
page 159 in palace
Let d be a metric on P(N) defined by d(A,B)=1/min(A+B).
Prove that (P(N),d) is compact.
Let K be the set of all functions f:N->R such that /\n 0<=f[n]<=1/n.
Recall that K is a compact set with respect to the sup metric.
For each AcN, define F.A:N->R by F.A(n) = 1/n * 1A(n).
Let E = {F.A : AcN}. E c K. Show that E is a closed subset of K.
Hence E is compact. Recall that (P(N),d) and (E,sup) are isometric.
Give an example of a metric space that is bounded but not totally bounded.
An infinite discrete metric space is bounded. When you consider balls of radius 1/7, you see that they are simply single-point sets. Hence the whole space is not a finite union of sets of diamater less than 1/7.
Give an example of a metric space that is bounded but not totally bounded. Don't use the discrete metric. Give an ambitious example.
Let X = { f:N->R : sup{|f(x)| : x:-N} <= 1 }.
Consider X as a metric space with the sup metric.
It is bounded because it is a closed ball.
X contains all characteristic functions of one-point subsets of N.
The distance between any two such functions equals 1.
Hence X is not a finite union of sets with diameter 1/2.
Give a definition of the Cantor set.
I = [0,1]. If J = [a,b]
let J(0) = [a , a+(b-a)/3],
let J(1) = [b-(b-a)/3 , b].
If AcN and n:-N, let I(A(n)) = I(1A(1), 1A(2), ..., 1A(n)).
If n:-N then let C[n] = U{I(A(n)) : AcN}.
Let's define C = //\\n:N [ C[n] ].
(Notice that C is an intersection of closed sets.)
Let (X,d) , (Y,g) be two metric spaces.
Consider f:X*Y->R.
(1) /\y:-Y f is continuous as a function of x:-X
(2) /\x:-X f is continuous as a function of y:-Y
Does f have to be continuous?
NO.
h(x,y) = min { |x|/|y|, |y|/|x| } for x!=0 and y!=0.
h(x,y) = 0 if x=0 or y=0.
f(0,0) = 0
f(x,y) = x*y / (x*x + y*y)
Let (X,d) , (Y,g) be two metric spaces.
Consider f:X*Y->R.
(1) /\y:-Y f is uniformly continuous as a function of x:-X
(2) /\x:-X f is uniformly continuous as a function of y:-Y
Does f have to be continuous?
NO.
h(x,y) = min { |x|/|y|, |y|/|x| } for x!=0 and y!=0.
h(x,y) = 0 if x=0 or y=0.
f(0,0) = 0
f(x,y) = x*y / (x*x + y*y)
Prove that the Cantor set has Lebesgue measure zero.
page 168 in the palace notebook