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

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Homework Help Overview

This discussion revolves around a topology problem concerning the properties of continuous maps and the relationship between open sets and their preimages. The original poster questions whether there is always an open set in the domain whose closure is contained within the preimage of an open set in the codomain.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of continuity and the properties of open and closed sets in topology. There is a distinction made between the closure of the preimage and the preimage of the closure of an open set. Some participants question the implications of different topological properties, such as second countability and local compactness.

Discussion Status

The discussion is ongoing, with participants offering insights into the nature of open sets and their preimages. Some guidance is provided regarding the use of metric spaces and open balls, but no consensus has been reached on the original question posed by the poster.

Contextual Notes

Participants note that the properties of the topology in question may influence the outcome, particularly regarding whether any open set contains a proper open subset. The implications of specific topological conditions, such as being second countable, locally compact, and Hausdorff, are also under consideration.

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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

 
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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?
 
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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.
 
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.
 
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What if my topology is second countable, locally compact and Hausdorff?
 
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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.
 
I see. Thanks.
 

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