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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!

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  • $\begingroup$ An open set in $Y$ is a subset of an open set in $T$. So if open sets in $Y$ cover $S$... $\endgroup$ – WimC Nov 21 '12 at 5:58
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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 $S$.

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