Definitions of relatively compact

Question

Let $X$ be a topological space and let $A \subseteq X$.

Definition 1. $A$ is relatively compact in $X$ if the closure of $A$ in $X$ is a compact subspace.

The above is standard, but there are some prima facie weaker conditions that are sometimes useful. For example:

Definition 2: $A$ is relatively compact in $X$ if there exists a compact $K$ such that $A \subseteq K$ and $K$ is contained in the closure of $A$ in $X$.

Definition 2'. $A$ is relatively compact in $X$ if there exists a compact $K \subseteq X$ with $A \subseteq K$.

We could even not bother having an actual compact subspace:

Definition 3. $A$ is relatively compact in $X$ if every open cover of $X$ has a finite subset that covers $A$, i.e. given open $U_i \subseteq X$ ($i \in I$) such that $X = \bigcup_{i \in I} U_i$, there is a finite $I' \subseteq I$ such that $A \subseteq \bigcup_{i \in I'} U_i$.

Clearly, definition 1 implies definition 2, and definition 2 implies definition 3. Furthermore, definition 2 and 2' are equivalent because closed subspaces of compact spaces are compact. Compact subspaces of Hausdorff spaces are closed, so if $X$ is Hausdorff, then definitions 1 and 2 are equivalent.

Question. Does definition 3 imply definition 1 in general? What if $X$ is Hausdorff, or $A$ is open?

Definition 3 is closely related to exponentiability: $X$ is exponentiable if and only if for every open $V \subseteq X$ and every $x \in V$ there is an open $U \subseteq V$ such that $x \in U$ and $U$ is relatively compact in $V$ in the sense of definition 3. When $X$ is Hausdorff, $X$ is exponentiable if and only if $X$ is locally compact.

Answer 1

Without regularity, definition $3$ does not imply definition $1$ in Hausdorff spaces.

Take a compact $T_2$ space $X=(X,\tau)$ which contains an open subset $G\subseteq X$ for which $X\setminus G$ is infinite and has empty interior.

The family $$\mathcal{B}=\{\{x\}\cup (U\cap G)\mid x\in U\in\tau\}\cup\tau$$ is a base for a topology $\sigma$ on $X$. This topology is evidently not regular, but being larger than $\tau$, it is Hausdorff. Since $\sigma$ is Hausdorff but not regular, it cannot be compact. Note that $G$ inherits the same subspace topology from both $(X,\tau)$ and $(X,\sigma)$.

Now, any subset of $G$ has identical closures in both $(X,\tau)$ and $(X,\sigma)$. In particular, since $G$ is dense in $(X,\tau)$, it is also dense in $(X,\sigma)$. Since $(X,\sigma)$ is not compact, $G$ is not relatively compact in $(X,\sigma)$ as per definition $1$.

On the other hand, suppose that $\mathcal{U}$ is a $\sigma$-open cover of $X$. For each $U\in\mathcal{U}$ let $V_U\subseteq X$ be a $\tau$-open set such that $U\subseteq V_U$ and $V_U\cap G=U\cap G$. Then $\mathcal{V}=\{V_U\mid U\in\mathcal{U}\}$ is a $\tau$-open cover of $X$. Since $(X,\tau)$ is compact, $\mathcal{V}$ contains a finite subfamily $\mathcal{V}'$ which covers $X$.

Put $\mathcal{U}'=\{U\in\mathcal{U}\mid V_U\in\mathcal{V}'\}$. This family might not cover $X$, but if $x\in G$ is contained in $V_U\in\mathcal{V}'$, then $x\in V_U\cap G=U\cap G$, and hence $x\in U\in\mathcal{U}'$. It follows that $\mathcal{U}'$ is a finite subfamily of $\mathcal{U}$ which covers $G$.

We conclude that $G$ is relatively compact in $(X,\sigma)$ as per definition $3$.

To see that this is nontrivial we can take $G=\mathbb{N}$ in $X=\beta\mathbb{N}$, or we can take $X=[0,1]$ and let $G$ be the complement of an infinite convergent sequence.

On the other hand, in the presence of regularity, we can establish that definition $3$ implies definition $1$ for Hausdorff spaces.

First observe that for $A\subseteq X$, under definition $3$, $A$ is relatively compact in $X$ if and only if it is relatively compact in $\overline A$. Thus the claim reduces to the statement that for a regular Hausdorff space $X$, a dense subset $A\subseteq X$ is relatively compact in the sense of definition $3$ if and only if $X$ is compact.

Now, recall ([1, Exercise 3.12.5, pg.222]) that a Hausdorff space is H-closed if it is a closed subspace of any other Hausdorff space in which it embeds. We have the following

1. A regular Hausdorff space is H-closed if and only if it is compact. Thus to prove the claim it is sufficient to show that $X$ is H-closed.
2. A Hausdorff space is H-closed if and only if any open cover $\mathcal{U}$ of it contains a finite subfamily $\mathcal{V}\subseteq\mathcal{U}$ whose union is dense. Thus if $A\subseteq X$ is dense and relatively compact (def.3), then $X$ is H-closed.

Putting these facts together yields the following.

If $X$ is a regular Hausdorff space, then $A\subseteq X$ is relatively compact in the sense of definition $3$ if and only if $\overline A$ is compact. $\quad\square$

[1]: Ryzard Engelking, General Topology, Revised Edition, Helderman Verlag Berlin, (1989).

Answer 2

Definition 3 does not seem to imply Definition 1 in general. Consider $\mathbb{N}$ with the topology with basis consisting of $\{0\}$ and $\{0,n\}$ for all $n\geq 1$. In this topology the singleton $A=\{0\}$ satisfies definition 3. However the closure of $A$ is the entire space $\mathbb{N}$ which is not compact (as witnessed by the open cover comprising of our basis elements). Note also that $A$ is open so that addresses another part of you question.

Answer 3

Great answer from Tyrone. I just want to complement it by stating the more general result:

If $X$ is a regular space, definitions 1, 2, 3 are equivalent.

This hold even if $X$ is not Hausdorff. A pretty straighforward proof is given in [C], Proposition 4.4, where a subset of $X$ relatively compact according to Def 1 is called precompact in $X$ and relatively compact according to Def 3 is called relatively compact to $X$.

The notion of relatively compact according to Def 3 (as well as other "relative" topological properties) was apparently first studied systematically by Arhangel'skii and collaborators, and the equivalence above is shown in [A]. [C] also shows that the space being a countable union of Def 3 sets is equivalent to a certain topological property that can be expressed in terms of Menger games. So there is an interest in that property, even without any separation axiom.

[A]: A.V. Arhangel’skii, From classic topological invariants to relative topological properties, Sci. Math. Jpn. 24 (2002), no. 1, 153-201 (I don't have access to this unfortunately, does not seem to be available online)

[C]: Steven Clontz, Applications of limited information strategies in Menger’s game, Commentat. Math. Univ. Carol. 58, No. 2, 225-239 (2017)

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One question I am wondering about:

Without assuming regularity, does Def 3 imply Def 2?

That is, is there an example of a Def 3 relatively compact subset $A$ of a space $X$ that is not contained in a compact subset of $X$?

Answer 4

Summarizing a few things:

Tyrone constructed a space that is D3 but not D1 (and possibly not D2).

Patrick points out that in regular spaces, D1, D2, D3 are all equivalent.

In the comments of Patrick's answer, Caruvana points out a Hausdorff example with a D3 but not D2 subset (if $[-n,n]\setminus\mathbb Q$ was D2, the space would be $\sigma$-compact).

It was noted in the question that D2 and D2' are always equivalent. Here are the details: D2 implies D2' directly. So assuming D2', we have $A\subseteq K$ compact. Then let $L=K\cap\overline{A}$: this set is closed in $K$, and since $K$ is compact, $L$ is also compact. Then $A\subseteq L\subseteq \overline{A}$, demonstrating D2.

It was noted in a comment to the question that D1' (the set is contained in a closed and compact set) is equivalent to D1. To see this, note D1 implies D1' since $\overline{A}$ would be closed and compact. Given D1', $A\subseteq K$ closed and compact implies $A\subseteq\overline{A}\subseteq K$ since $K$ is closed. Then $\overline{A}$ is a closed subset of compact $K$ and is thus compact.

Assuming Hausdorff (or just KC), D1 and D2 are equivalent. D1 implies D2 directly, and given D2 and Hausdorff, $L\subseteq \overline{A}$ means, since $L$ is compact and therefore closed, $L=\overline{A}$ is compact.

Here's an example of a $T_0$ space with a D2 but not D1 subset: let $X=\omega$ with basic open sets of the form $[0,n)$ for $n<\omega$. Then $\{0\}$ is compact and therefore D2, but not D1, since $\overline{\{0\}}=X$ is not compact.

With a bit more thought, here's a $T_1$ example. Let $X$ be an infinite compact $T_1$ space, and let $Y$ be an infinite set. Give $Z=X\cup Y$ the topology where points of $X$ have basic open sets given by the original topology of $X$, and let points $y$ of $Y$ have basic open sets of the form $\{y\}\cup X\setminus F$ where $F$ is a finite (and therefore closed) subset of $X$. Then $Z$ is $T_1$ with a compact and therefore D2 subset $X$. But $\overline{X}=Z$, which contains an infinite closed discrete set $Y$, and therefore is not compact. Thus $X$ is D2 but not D1.