Page 154 - DMTH503_TOPOLOGY

Topology

Notes Clearly, A X
We claim A  X.


Suppose not, then X A   ...(3)


Let y  X A be arbitrary. A is closed and hence X A is open. It amounts to saying that
y  X A  T.

By definition of base


T
y  X A    B  s.t. y B  X A.


y
y
In particular 
x  X A T


n o
  B n  s.t. x  X A.

o o n

Now x  X A  x   A  A
n o n o
 x   A ...(4)
n o
x  B  x  A, according to (1) and (2), Contrary to (4).
n o n n o
Hence our assumption X A   is wrong.
Consequently X A   i.e. X  A

Thus, we have shown that
 A  X s.t. A  X and X is enumerable set. By definition, this proves that X is separable.
Theorem 8: Every second axion space is hereditarily separable.

Proof: Let (Y, ) be a subspace of second axion, i.e. second countable space (X, T).
To prove the required result, we have to show that (Y, ) is second countable and separable since
every second countable space is separable. [Refer theorem (7)].

Now it remains to show that (Y, ) is second countable. Now write the proof of Theorem (6).

Example 3: Prove that (R, ) is a second axiom space (Second countable.).

Solution: We know that Q is a countable subset of R. If we write
 = {(a, b) : a < b and a, b  Q}
Then  forms a countable base for the usual topology and R so that R is second countable.

Example 4: Prove that (R , ) is second countable.
2
Solution: If we write
 = {S (x) : x, r Q}
r
2
2
then  forms a countable base for the usual topology on R . Hence (R , ) is second countable
space.

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