Page 114 - DMTH503_TOPOLOGY

Topology

Notes
 B  A C
A B =   A  C

C
C
C
C
 A B  B A = (B A) = X = 
i.e., A B = 
Thus X can be represented as the union of two disjoint non-empty closed sets.
This contradicts the given hypothesis (iii) and thus (iv) must be true.
(iv)  (i)

Let (iv) be true
Suppose that X is disconnected.
Then there exist disjoint non-empty open sets G and H such that X = G H.

Since G and H are open and G H = , is follows G H =  and G H = .

This contradicts the given hypothesis (iv) and thus (i) must be true.
Hence the proof of the theorem.
Theorem 2: The closure of a connected set is connected
OR

If A is connected subset then show that A is also connected.
Proof: Let (X, T) be a topological space and A be a subset of X.

If A is connected, then we have to show that A is also connected.

If A is not connected then it has a separation.

Let A = G | H
So by theorem, Let (X, T) be a topological space and let E be a connected subset of (X, T). If E has
a separation X = A | B, then either E  A or E  B, we have

A  G or A  H

If A  G

 A  G

 A  G

 A H  G H

 A H   ( G and H are separated.) ...(1)

Also A = G H ...(2)

 H  A

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