Page 80 - DMTH503_TOPOLOGY

Topology

Notes Evidently
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f (B)  f (B) …(1)
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  f f (B)    f f (B) 
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    f f (B)       f f (B)    
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    f f (B)    f f (B)  (on using (1))
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 ff (B)  f f (B) 
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 f (B)  f (B)
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 f is continuous.
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Similarly we can show that if f is open, that f is continuous
Theorem 6: A map f : (X, T)  (Y, V) is closed iff
f(A)  f(A) for every A  X.
Proof: Let (X, T)  (Y, V) be closed map and A  X arbitrary.
To prove f(A)  f(A)
A is closed subset of X, f is closed.
 f(A) is closed subset of Y.
 f(A) f(A) …(1)

But A  A

 f(A)  f(A)

 f(A)  f(A)

 f(A)  f(A)  f(A), By (1)

 f(A)  f(A),

Conversely, suppose f(A)  f(A)  A  X. …(2)

To prove that f is closed.

Let F be a closed subset of X so that F  F

F  F  f(F)  f(F) …(3)

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