Prove that a function between metric spaces is continuous if and only if the pre

Question

Prove that a function between metric spaces is continuous if and only if the preimage of every closed set is a closed set.

More precisely, this means: a function is continuous if and only if the preimage of every closed subset of the codomain is a closed subset of the domain.

Hint: use the fact that a function is continuous if and only the pre-image of every open subset of the codomain is open in the domain, and use the definition of closed set,

and use the fact that forming preimage is compatible with complements.

Explanation / Answer

NOTE : The following are equivalent for a function f : X Y between two metric spaces: (i) f is continuous (ii) Whenever V is an open subset of Y , the inverse image f1 (V ) is an open set in X

USING THE NOTE, WE WILL SOLVE THE QUESTION

for a function f : X Y between two metric spaces: f is continuous if and only if whenever F is a closed subset of Y , the inverse image f1(F) is a closed set in X

proof :: Assume that f is continuous and that F Y is closed. Then Fc is open, and , f1 (Fc ) is open. Since inverse images commute with complements, (f1 (F))c = f1 (Fc ). This means that f1 (F) has an open complement and hence is closed.

now, Assume that the inverse images of closed sets are closed. it suffices to show that the inverse image of any open set V is open. But if V is open, the complement, Vc  is closed, and hence by assumption f1 (Vc ) is closed. Since inverse images commute with complements, (f1 (V ))c = f1 (Vc ). This means that the complement of f1 (V ) is closed, and hence f1 (V ) is open

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