Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have shown in class that if $E$ is a compact subset of a metric space $X$, then $\forall Y \subset X$ such that $E \subset Y$, E is compact in Y with the same metric as $X$. That is, compactness in a metric space is generalizable to any space with the same metric.

I'm trying to convince myself of this same notion in a general topological space. Take $T$ as a topological space, and $S \subset T$ a compact set. If we construct $Y \subset T$ with the induced topology from $T$, does it follow that if $S \subset Y$ then $S$ is compact in Y?

Any help would be greatly appreciated. Many thanks!

share|cite|improve this question
An open set in $Y$ is a subset of an open set in $T$. So if open sets in $Y$ cover $S$... – WimC Nov 21 '12 at 5:58
up vote 1 down vote accepted

Yes, it does: the compactness of $S$ depends only on the topology on $S$.

Let $\langle T,\tau\rangle$ be a topological space and $S$ a compact subset of $T$. Let $\tau_S$ be the subspace topology on $S$. Let $\mathscr{U}\subseteq\tau_S$ be an $S$-open cover of $S$. For each $U\in\mathscr{U}$ there is a $V_U\in\tau$ such that $U=S\cap V_U$, so $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$ is a $T$-open cover of $S$. $S$ is compact in $T$, so there is a finite $\mathscr{V}_0\subseteq\mathscr{V}$ that covers $S$. Let $\mathscr{U}_0=\{U\in\mathscr{U}:V_U\in\mathscr{V}_0\}$; then $\mathscr{U}_0$ is a finite subfamily of $\mathscr{U}$ covering $S$. In other words, the space $\langle S,\tau_S\rangle$ is a compact space in its own right, without any reference to $T$.

Now suppose that $S\subseteq Y\subseteq T$. Let $\tau_Y$ be the relative topology on $Y$ as a subspace of $T$ and $\tau_S'$ the relative topology on $S$ as a subspace of $\langle Y,\tau_Y\rangle$. Then $\tau_S'=\tau_S$, i.e., $S$ has the same topology as a subspace of $Y$ as it has as a subspace of $T$. (This is a straightforward exercise; if you get stuck, just ask.) Thus, if $\mathscr{U}$ is a $\tau_Y$-open cover of $S$ in $Y$, $\{U\cap S:U\in\mathscr{U}\}$ is a $\tau_S$-open cover of $S$ in $S$. $S$ is compact, so $\{U\cap S:U\in\mathscr{U}\}$ has a finite subcover, and from that it’s easy to extract a finite subfamily of $\mathscr{U}$ that covers $S$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.