Page 81 - DMTH503_TOPOLOGY

Unit 7: Continuous Functions

Notes
Also, by (2), f(F)  f(F)
Combining this with (3),

f(F)  f(F)

But f(F)  f(F) [For C  C is true for any set C]
Combining the last two.

f(F)  f(F).

 f(F) is closed.
Thus F is closed.  f(F) is closed.

 f is closed map.
Theorem 7: A function f : (X, T)  (Y, V) is continuous iff

1

B
 f (B)  f  1  , B  Y.
 

1

or f (B )    f (B) 

1

Proof: Let f : (X, T)  (Y, V) be a topological map. Let B  Y be arbitrary.
(i) Suppose f is continuous.

1
1


To prove that f (B)  f (B )




B  Y  B° is open in Y.
 f (B°) is open in X. For f is continuous.
-1



1
1
  f (B )  f (B ) …(1)


 
B°  B  f (B°)  f (B)
-1
-1
-1
-1
 f (B)  f (B°)
1 
1
1 

 f (B)      f (B )  f (B ), [by (1)]








1
1


 f (B)    f (B )


Proved.
1

(ii) Suppose f (B)    f (B ) …(2)

1




To prove f is continuous.
Let G be an open subset of Y and hence G = G°
-1
If we show that f (G) in open in X, the result will follow:


1
1
  f (G)  f (G ), [by (2)]


–1
= f (G)
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