Are closed sets in a pre-image always contained in an open set in the domain?

[ehrenfest]

Homework Statement

This is a topology problem.
I have a continuous map from X to Y, and I take an open set U in Y, and I look at its preimage. Is it true that there must always be an open set in X whose closure is in the preimage of U?

I know that there is always an open set whose closure is in the preimage of the closure of U. But that is not the same thing...

Homework Equations

The Attempt at a Solution

 

[Dick]

The preimage of an open set in Y IS an open set in X. That's basically the definition of continuous function. If it's closure is also the preimage, then the answer is no. Not every open set is also closed. Why are you muddling this up?

 

[ehrenfest]

I am not asking whether the closure of the preimage of an open set is in the preimage, I am asking if there is any open set whose closure is in the preimage.

 

[Dick]

Depends on whether your topology guarantees that any open set contains a proper open subset. The preimage is after all, just an open set. It has no special properties beyond that.

 

[ehrenfest]

What if my topology is second countable, locally compact and Hausdorff?

 

[Dick]

If it's a metric space just find an open ball in the set. Call it's radius r, and take the ball of radius r/2 as your set.

 

[ehrenfest]

I see. Thanks.