Page 94 - DMTH503_TOPOLOGY

Topology

Notes  S F =  [ x F]
r (x)
 S X – F
r (x)
 S F.
r (x)
Given x F, any open sphere S s.t.
r(x)
S  F
r(x)
By definition, this proves that Fis open.
Conversely suppose that F is open in X.
To prove that F is closed in X.
Let x F be arbitrary, then x F.
Fis open,  R s.t., S F
+
r r(x)
 S F = 
r (x)
 (S – {x}) F = 
r (x)
 x D (F).
Thus, any x F  x D (F)
i.e. any x X – F  x X – D (F)
 X – F X – D (F) or D (F) F
 F is closed.
Theorem 3: In any metric space (X, d), each open sphere is an open set.

Proof: Let (X, d) be a metric space. Let S 0 r (x ) be an open sphere in X.
0
To prove that S 0 r (x ) is an open set.
0
Let x  S 0 r (x ) be arbitrary, then d(x, x ) < y 0
0
0
Write r = r – d (x, x ) ...(1)
0 0
By definition S 0 r (x ) = {y X : d (y, x ) < r }
0
0
0
S = {y X : d (y, x) < r}.
r (x)
We claim S  S r (0) (x )
r (x) 0
Let y S be arbitrary
r (x)
Then d (x, y) < r
d (y, x )  d (y, x) + d (x, x )
0 0
< r + d (x, x ) = r . [on using (1)]
0 0
 d (y, x ) < r
0 0
 y  S
0 r (x ) )
0
and y S  y  S
r (x) 0 r (x )
0
 S  S
r (x) 0 r (x )
0
Thus we have shown that for given any x  S 0 r (x ) , r > 0 s.t. S  S 0 r (x ) .
r(x)
0
0
By definition, this proves that S 0 r (x ) is an open set.
0

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