Prove that closed subsets of a locally compact space are locally compact.

Closed subsets of a locally compact space are locally compact because for every point in the closed subset, there exists a neighborhood contained in a compact set. This is established through the intersection of compact closures with the closed subset. Therefore, the property of local compactness is preserved in closed subsets.
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Answered by Anonymous | 2025-07-04

To prove that closed subsets of a locally compact space are locally compact, let's walk through the definitions and the proof in a clear and systematic way:

Definitions :

A locally compact space is a topological space in which every point has a neighborhood base consisting of compact sets. This means for every point in the space, there is some neighborhood around that point that is compact (i.e., closed and bounded in Euclidean space).

A closed subset is a subset such that its complement is open in the space.



Proof Outline :
Consider a locally compact topological space X and let C be a closed subset of X .

Take any point x in C . Since X is locally compact, there exists an open set U in X containing x such that the closure U is compact.

Note that U ∩ C is the intersection of a compact set and a closed set, which is itself compact.

The set U ∩ C is an open neighborhood of x in C . Note the closure of U ∩ C in C is U ∩ C , which we have already concluded to be compact.

Therefore, every point x in C has a neighborhood in C with a compact closure, satisfying the requirement for C to be locally compact.


Hence, closed subsets of a locally compact space are indeed locally compact.


In summary, one can see that by using properties of compactness and closed sets, the proof is constructed to show that the locally compact nature transfers from the space to its closed subsets. This is a fundamental property useful in various areas of topology.

Answered by LiamAlexanderSmith | 2025-07-06