Page 78 - DMTH503_TOPOLOGY

Topology

Notes i.e., f (Y)  f (G) is closed in X,
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i.e., X  f (G) is closed in X,
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i.e. f (G) is open in X,
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 any G  Y is open  f (G) is open in X
This proves that f is continuous map.
Theorem 3: Let f : (X, T)  (Y, U) be a map, Let  be a sub-base for the topology U on Y. Then f is
continuous iff f (S) is open in X whenever S  
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or
f is continuous  the inverse image of each sub-basic open set is open.
Proof: Let f : (X, T)  (Y, U) be continuous map. Let  be a sub-base for the topology U on Y. Let
S   be arbitrary.
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To prove that f (S) is open in X.
S    S  U ( S  U  S is open in Y)
 f (S) is open in X, (by Theorem 1).
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Conversely, suppose that f : (X, T)  (Y, U) is a map such that f (S) is open in X whenever S  ,
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 being a sub-base for the topology U on Y. Let B be a base for U on Y.
To prove that f is continuous.

Let G  Y be an open set, then G  U.
By definition of base,
G  U      s.t. G =  {B : B   } …(1)
1 1
By the definition of sub-base, any B   can be expressed as

n
B  S i for same choice of S , S , … S  
i 1 1 2 n

1

f (B) f  1  n S   n f 1    …(2)

S
  i 1 i   i 1 i

n  1
S
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By hypothesis, f (S ) is open in X, Being a finite intersection of open sets in X, f   is open
i
i
i 1

in X, i.e. f (B) is open in X
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
1
i.e. f (G) = f [ {B : B  1 }]
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1

=    f (B) : B  1  
An arbitrary union subsets of X
open subset of X.
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 f (G) is open in X.
Thus we have shown that
any G  Y  f (G) is open in X.
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This proves that f is continuous.
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