Page 100 - DMTH503_TOPOLOGY

Topology

Notes A is closed X – A is open.
By definition of open set,
+

any x X – A   R s.t. S  X A S  A  
r r (x) r (x)
 d (X, A) r d (x, A) 0.
Conversely let A be any subset of a metric space (X, d).
Let any x X – A  d (x, A) 0.
To prove that A is closed.
Let x X – A be arbitrary so that, by assumption


d (x, A) = r 0  S  A    S r  X A
r
(x) (x)
 x X – A   r  R s.t. S r  X A.
+

( x)
By definition, this implies X – A is open
 A is closed
Proved.

Problem: In any metric space, show that

X A = (X – A)°
or (A) = (A ) . 

Solution: (A) = X A
= X – Intersection of all closed super sets of A

= X  F where F is closed and F A
i i i
i


= (X F ) where X – F is open and X – F X – A
i
i
i
i
= Union of open subsets of X – A = A
= (A ) . 
Proved.
Problem: In any metric space (X, d), prove that A is open A° = A.
Solution: Let A be a subset of a metric space (X, d). By definition of interior,

A° =  {S r (x) : S  A} ...(1)
r
(x)
since every open sphere is an open set and arbitrary union of open sets is open.
Consequently,
A° is an open set. ...(2)

By (1), it is clear that A° A ...(3)
and A° is largest open subset of A. ...(4)

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