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Let (X,d) be a bounded metric space.

Let H be the set of all closed subsets of X except the empty set.

Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

(H,h) is the Hausdorff metric space.

Prove that d(a,b) = h({a},{b}).

Let H be the set of all closed subsets of X except the empty set.

Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.

(H,h) is the Hausdorff metric space.

Prove that d(a,b) = h({a},{b}).

page 179 in the palace notebook

Let (X,d) be a metric space. f:X->X.

Suppose that f satisfies:

/\x,y:-X [ x!=y ==> d(f(x),f(y)) < d(x,y) ].

Decide if we can conclude that

\/0<K<1 /\x,y:-X d(f(x),f(y)) <= K*d(x,y).

Suppose that f satisfies:

/\x,y:-X [ x!=y ==> d(f(x),f(y)) < d(x,y) ].

Decide if we can conclude that

\/0<K<1 /\x,y:-X d(f(x),f(y)) <= K*d(x,y).

NO.

Consider [2,oo) with the Euclidean metric.

Put f(x) = x + 1/x, for x>=2. Notice that f(x)>=2.

To show that f satisfies the desired condition,

use the fact that f is increasing.

Since f has no fixed point, it cannot be a contraction.

Consider [2,oo) with the Euclidean metric.

Put f(x) = x + 1/x, for x>=2. Notice that f(x)>=2.

To show that f satisfies the desired condition,

use the fact that f is increasing.

Since f has no fixed point, it cannot be a contraction.

Prove that there exists a continuous function f:[0,1]->R that
is neither increasing nor decreasing on any subinterval of [0,1].

Consider the set of all continuous f:[0,1]->R. Equip it with
the sup metric. It's a complete metric space. Let I be a
subinterval of [0,1]. Let A(I) denote the set of all continuous
f:[0,1]->R that are increasing on I. (x<y==>f(x)<=f(y)). A(I)
is closed. Let B(I) be the set of all continuous f:[0,1]->R
that are decreasing on I. B(I) is closed. Let K = A(I) u B(I).
K is closed and Int(K) = 0. Let {I[n]} be a sequenc

Let z be a complex number such that |z|<1. Let m be a natural
number.

Express 1 / (1-z)^m as the sum of a power series.

Express 1 / (1-z)^m as the sum of a power series.

1 / (1-z)^m = +(n=0 to oo) [ z^n * (n+m-1)! / (n!*(m-1)!) ]

page 190 in the palace notebook

page 190 in the palace notebook

Let z be a complex number.

z * conjugate(z) = ?

z * conjugate(z) = ?

z * conjugate(z) = |z|*|z|

Let z be a complex number other than zero.

conjugate(z) / |z|^2 = ?

conjugate(z) / |z|^2 = ?

conjugate(z) / |z|^2 = 1/z

Let z be a complex number other than zero.

z / |z|^2 = ?

z / |z|^2 = ?

z / |z|^2 = 1/conjugate(z)

Let h : |C- > |C be defined h(z)=|z|*|z|.

Where is h diffable?

Where is h diffable?

Only at z = 0.

Show a complex function that is diffable only at one point.

f(z)=|z|^2

f(x,y)=x^2+y^2

f is diffable only at zero

f(x,y)=x^2+y^2

f is diffable only at zero

Consider a complex function defined on an open subset of |C.
Suppose that it is diffable at some point in that set. Does it
have to be diffable in a neighborhood of the point?

NO.

There are complex functions that are diffable only at one point.

1) z->|z|*|z| at 0

2) f(x,y)=xy at 0

There are complex functions that are diffable only at one point.

1) z->|z|*|z| at 0

2) f(x,y)=xy at 0