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Let $X$ be a locally compact topological space, and let $A$ be a subset of $X$ such that $A$ is also closed in $X$. Then $A$ is also locally compact when regarded as a subspace of $X$.

Munkres' Proof:

Let $x \in A$. We need to show that $A$ is locally compact at $x$. Since $X$ is locally compact and $x \in X$, therefore we can find a compact subspace $C$ of $X$ and an open set $U$ of $X$ such that $$x \in U \subset C. $$ Then we also have $$ x \in A \cap U \subset A \cap C. $$ Now as $U$ is open in $X$, so $A \cap U$ is open in the subspace $A$ of $X$. And, as $A$ is closed in $X$, so $C \cap A = A \cap C$ is closed in the subspace $C$ of $X$, and since $C$ is compact, therefore $A \cap C$ is also compact when regarded as a subspace of $C$.

How to show from here that $A \cap C$ is also compact when regarded as a subspace of $A$?.

And, last but not the least, what is the most general result of this type that one can come up with?

Saaqib Mahmood
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  • $Y \subset X $ . Then $A\subset Y$ is $\tau_Y$-compact iff $A$ is $\tau_X$-compact. See https://math.stackexchange.com/a/3440331/977780 – Sourav Ghosh Aug 08 '22 at 13:30
  • Your concluding question does not make much sense to me; what is the set of "types" under consideration? – Lee Mosher Aug 08 '22 at 13:37

1 Answers1

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Compactness of a topological space is an intrinsic property of the topology on that space (every cover of the space by open subsets has a finite subcover). So to say that $A \cap C$ is compact "when regarded as a subspace of $A$" means that $A \cap C$ is compact with respect to the subspace topology relative to $A$: every cover of $A \cap C$ by subsets of $A \cap C$ that are open relative to $A$ has a finite subcover.

Now we have $A \cap C \subset A \subset X$, and it is an exercise in the subspace topology to prove that if $P \subset Q \subset R$ and if $R$ is a topological space then the subspace topology on $P$ relative to $R$ is equal to the subspace topology on $P$ relative to (the subspace topology on $Q$ relative to $R$).

Therefore, compactness of $A \cap C$ is well-defined whether you use the subspace topology relative to $A$ or to $X$.

Lee Mosher
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