Page 236 - DMTH503_TOPOLOGY

Topology

Notes
Example 2: If a metric space (X, d) is totally bounded, then X is bounded.
Solution: Let (X, d) be a totally bounded metric space so that it contains an  - net for every  > 0.
Let A  X be an  - net then:
(i) A is finite
(ii) X = {S (a) : a  A}

(i) A is bounded  d(A) is finite.
(ii) d(X)  d(A) + 2 = a finite quantity.
d(X)  a finite quantity
X is a bounded set. Hence proved.

Example 3: Every totally bounded metric space is separable.

Solution: Let (X, d) be totally bounded metric space so that X contains an  - net A   > 0.
n n
To prove that X is separable.
A is  - net  A is finite and X = {S(a,  ) : a  A }.
n n n n
Write A = {A : n  N}
n
Being an enumerable union of finite sets A is enumerable.
A  X  A  X  X  A  X. ...(1)
Let x  X be arbitrary and let G be an open set s.t. x  G.
By definition of open set

G  S ...(2)
(x,  ) n
Also A S  . For A is  - net.
n (x, e n ) n
This S (x, e n ) A  

 G A   [by (2)]

 x  A
 Any x  X  x  A

Consequently X  A

In view of (1), this X = A
This leads to the conclusion that X is separable.
Theorem 4: Lebesgue covering lemma: Every open cover of sequentially compact metric space has
a Lebesgue number.
Proof: Let {G : i  } be an open cover for a metric space (X, d). A real number  > 0 is called a
i
Lebesgue number for the cover if any A  X s.t. d(A) <   A  G for at least one index i  .
i 0 0
Let {G : i  } be an open cover of a sequentially compact metric space (X, d).
i
To prove that the cover {G } has a Lebesgue number.
i i  

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