Continuity of a Function with Inverse Preimage Condition

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Homework Statement

Suppose f:X-->Y
suppose for each open set U in Y s.t U contains some element f(x), we have f^(-1)(U) is open in X.
Does this imply f is continuous

Homework Equations

U is not quite an arbitrary open set of Y since there could be an open set of Y that does not interesct the image f(X).
Suppose V is such a set.
Then can we say f^(-1)(V)= the empty set? Or is it just undefined?

This isn't a homework question, just curious.

The Attempt at a Solution

I believe it is continuous since, otherwise the proof I'm reading in a textbook wouldn't work (they just ommited this part).

Could anyone explain the reasoning for this?
thank you

 

[n!kofeyn]

You're right since if the open set U didn't contain the image of any point in X, then f^-1(U) would be empty (since nothing in X was mapped to U), and the empty set is an open set. So they are just excluding that case since it doesn't really carry any information with it.

Also, this version of continuity is equivalent to the epsilon-delta definition of continuity, as well as a few other versions. For example, if for every closed set in Y, f inverse of that closed set is closed in X, then f is continuous. You can prove it from the above statement by taking complements.