Page 136 - DMTH503_TOPOLOGY

Topology

Notes Now if  is a fanily of finite numbr of sets in , say
*
 = { ] –n , n [, ] –n , n [, …, ] –n , n [ }
*
1 1 2 2 K K
*
and if n = max {n , n , …, n }, then
1 2 K
K
n *   ] n ,n [ 
i
i
 i 1
Thus it follows that no finite sub-family of  cover .
Hence (, U) is not compact.
Theorem 9: A closed and bounded subset (subspace) of  is compact.
Proof: Let I = [a , b ] be a closed and bounded subset of . Let G = {(c , d ) : i  } be an open
1 1 1 i i
covering of I .
1
To prove that  finite subcover of the original cover G.
Suppose the contrary.
Then  no finite subcover of the cover G.

Divide I into two equal closed intervals.
1
 a  b 1   a  b 1 
1
1
 a ,  and  , b 1  .
1
 2   2 
Then, by assumption, at least one of these two intervals will not be covered by any finite
subclass of the cover G. Call that interval by the name I .
2
Write I = [a , b ]
2 2 2
 a  b 1   a  b 1 
1
1
Then [a , b ] = a , or , b 1  .
2 2  1  
 2   2 
 a  b 2   a  b 
Divide I into two equal closed intervals a , 2  and  2 2 , b 2  . Again by assumption, at

2
2  2   2 
least one of these two intervals will not be covered by any finite sub-family of the cover G. Call
that interval by the name I .
3
Write I = [a , b ].
3 3 3
Repeating this process an infinite number of times, we get a sequence of intervals I , I , I , … with
1 2 3
the properties.
(i) I  I  n  N.
n n + 1
(ii) I is closed  n  N.
n
(iii) I is not covered by any finite sub-family of G.
n
(iv) lt [I ] = 0, where |I | denotes the length of the interval I and similar is the meaning of
n  n n n
|[a , b ]|.
r r
Evidently the sequence of intervals I  satisfies all the conditions of nested closed interval
n
property.

This  I  
n
n 1


So that  a number p  I .
0 n
n 1

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