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Consider the function $f(x)=x+2$. Let the domain be $\left(0,1\right)$, then the range is $\left(2,3\right)$. $\left(0,1\right)$ is not compact in $\mathbb{R}$ since it is not closed in $\mathbb{R}$. But if I consider the subspace topology with $\left(0,1\right)$, $\left(0,1\right)$ is closed (in $\left(0,1\right)$). But since $\left(2,3\right)$ is not compact in $\mathbb{R}$ and $f$ is continuous, $\left(0,1\right)$ should not be compact even with the subspace topology. I was wondering whether there is a general result like: if a set $S$ is not compact in $T$, then it is not compact in any set $U$ such that $S\subseteq U \subseteq T$. In other words, one cannot make a non-compact set compact by applying a subspace topology.

Ypbor
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    Yes, that general result is true. It follows by definition of compactness and definition of subspace topology. – geetha290krm Jan 11 '23 at 08:45
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    Use the fact that the inclusion is continuous and that images of compact sets are compact to prove to claim – IsAdisplayName Jan 11 '23 at 08:46
  • The theorem is, a subspace is $\mathbb R$ is compact if an only if it is closed and bounded in $\mathbb R.$ So the fact that $(0,1)$ is closed and bounded in the metric space $(0,1)$ does matter, and the theorem doesn't apply. "Closed and bounded" is not the definition of compact. – Thomas Andrews Jan 11 '23 at 08:46
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    Compactness is just about the topology. A space is not "compact in" a parent space, it is just either a compact space or not. But the theorem about which subspaces of $\mathbb R$ (or more generally, $\mathbb R^n,$) are compact is specifically about subspaces. – Thomas Andrews Jan 11 '23 at 08:51

1 Answers1

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The general result is true, and follows from the general definition of compactness.

Theorem. Let $S\subseteq U\subseteq T$. If any open cover of $S$ in $U$ has a finite subcover, then so does any open cover of $S$ in $T$.

Proof. For any collection of open sets $\{V_\alpha\}$ with $V_\alpha\subseteq T$ that covers $S$, it restricts to a collection $\{U\cap V_\alpha\}$ of open sets $U\cap V_\alpha\subseteq U$ that still covers $S$. Therefore, there exists a finite subcollection which covers $S$ by the hypothesis, and this will "lift" back to a finite subcollection of $\{V_\alpha\}$. $\square$

In particular, we can choose $U=S$. This shows that for any subset $S\subseteq T$, we have "$S$ is compact (in $S$)" if and only if "$S$ is compact in $T$" for any ambient space $T$. Hence, we see that compactness is an inherent notion to the topological space and does not depend on whether it is the subspace of some larger space. (Unlike e.g. the notion of open-ness or closed-ness.)

YiFan Tey
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