Page 134 - DMTH503_TOPOLOGY

Topology

Notes so that
d(x, A) – d(y, A) d(x, y).
The same inequality holds with x and y interchanged, continuity of the function d(x, A) follows.

Now we introduce the notion of Lebesgue number. Recall that the diameter of a bounded subset
A of a metric space (X, d) is the number
sup {d(a , a ) | a , a A}
1 2 1 2
Lemma (1) (The Lebesgue number Lemma): Let  be an open covering of the metric space (X, d). If
X is compact, there is a 0 such that for each subset of X having diameter less than , there exists
an element of  containing it.
The number  is called a Lebesgue number for the covering .

Proof: Let  be an open covering of X. If X itself is an element of , then any positive number is
a Lebesgue number of . So assume X is not an element of .
Choose a finite subcollection {A , ..., A } of  that covers X. For each i, set C = X – A , and define
1 n i i
f: X  be letting f(x) be the average of the numbers d(x, C ). That is,
i
1 n
f(x) =  (x, c )
n  i 1 i
We show that f(x) > 0 for all x. Given x  X, choose i so that x  A . Then choose  so
i
-neighborhood of x lies in A . Then d(x, c ) , so that f(x) /n.
i i
Since f is continuous, it has a minimum value  we show that  is our required Lebesgue
i
number. Let B be a subset of X of diameter less that . Choose a point x of B; then B lies in the
0
-neighborhood of x . Now
0
 f(x )  d(x , C ),
0 0 m
where d(x , C ) is the largest of the number d(x , C ). Then the -neighborhood of x is contained
0 m 0 i 0
in the element A – X – C of one covering .
m m
Definition: Uniformly Continuous
A function F from the metric space (X, d ) to the metric (Y, d ) is said to be uniformly continuous
X Y
if given  > 0, there is a  > 0 such that for every pair of points x , x of X,
0 1
d (x , x ) < d  d (f(x ), f(x )) < .
x 0 1 Y 0 1
Theorem 7: Uniform Continuity Theorem

Let f: X Y be a continuous map of the compact metric space (X, d ) to be metric space (Y, d ).
x y
Then f is uniformly continuous.
Proof: Given > 0, take the open covering of Y by balls B (y, /2) of radius /2. Let A be the
open covering of X by the inverse images of these balls under f. Choose to be a Lebesgue
number for the covering A. Then if x and x are two points of X such that dx(x , x ) < , the two
1 2 1 2
point set {x , x } has diameter less than . So that its image {f(x ), f(x )} lies in some ball B (y, /2).
1 2 1 2
Then dy (f(x ), f(x ) < , as desired.
1 2
Finally, we prove that the real numbers are uncountable. The interesting thing about this proof
is that it involves no algebra at all-no decimal or binary expansions of real numbers or the
like-just the other properties of .

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