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If $X$ is a (Hausdorff) topological space and $K_0\subset K_1\subset...\subset K_n\subset...\subset X$ are compact subsets such that $X=\bigcup_{n\in\mathbb{N}}K_n$, is it true that for any compact subset $K\subset X$ there is an $N\in\mathbb{N}$ such that $K\subset K_N$? Is it true if other conditions are added to $X$ (such as being normal, metrizable, locally compact or something else)?

I'm trying to generalize the behaviour of $X=\mathbb{R}^d$ (for $d\in\mathbb{N}^*$), where this is the case taking $K_n$ to be $[-n,n]^d$ for all $n\in\mathbb{N}$. If it isn't true for all sequences $(K_n)_{n\in\mathbb{N}}$ of a general $\sigma$-compact Hausdorff space $X$, is it at least true that there is such a sequence of compact sets with that property?

Is there a problem if we change countable sequences of compacts sets by other unions indexed by larger (possibly uncountable) ordinals?

Thanks in advance to anyone who can give any insight.

Susana Santoyo
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  • I think it's false for the space $\mathbb Q$ of rational numbers. – user14111 Jan 11 '24 at 02:54
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    Of course it fails if $X=\mathbb Q$ and $(K_n)$ is a sequence of finite sets. More generally, suppose $(K_n)$ is any sequence of compact subsets of $\mathbb Q$. Choose $x_n\in(\mathbb Q\cap(0,\frac1n))\setminus(K_0\cup\cdots\cup K_n)$ and let $K={0,x_1,x_2,\dots}$. – user14111 Jan 11 '24 at 02:59

2 Answers2

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Firstly, this is not generally true even in $\mathbb R$. Take $K_n=[-n,-\frac{1}{n}]\cup \{0\} \cup [\frac{1}{n},n]$, then $K=[-1,1]$ is compact, but will not be contained in any $K_n$.

Secondly, $\mathbb Q$ does not even have any such sequence of compact sets. To see this, observe that no compact set has any interior point, so if $K_1\subset\dots \subset K_n\subset \dots$ is a sequence of compact sets, then let $x_0=0$ and choose $x_n\in [-\frac{1}{n},\frac{1}{n}]\backslash K_n$. Then $K=\{x_n\mid n\in \mathbb N\}$ is compact and not contained in any $K_n$.

Remark.

You can generalize the situation in $\mathbb R^n$, however. If $X$ is "exhaustible by compacts", which is equivalent to being $\sigma$-compact and every point having a compact neighborhood ("weakly locally compact"), then there is indeed such a sequence of compact sets. So $\sigma$-compact, locally compact Hausdorff is more than sufficient.

Finally, allowing uncountable sequences of compacts doesn't really let you drop the $\sigma$-finite condition, as you might hope - for example, an uncountable discrete space has no chain of compact sets whose union is the entire space, since every chain of compact sets must be countable (the only compact sets are finite).

Edit: important points from the comments.

As bof points out, there are some cases where allowing uncountable chains will give you an exhaustion by compacts, e.g., the ordinal space $[0,\omega_1)$ with the order topology. I don't know of a general criterion for when a space does or does not have such an exhaustion.

Also, to expand on user14111's comment, since weak local compactness and $\sigma$-compactness gives an exhaustion such that $K_n\subseteq \operatorname{int}(K_{n+1})$ (see the above link for references to a proof of this), the sets $\operatorname{int}(K_{n})$ are an open cover of any compact subset $K$, hence they have a finite subcover. Taking the largest $n$ in this subcover gives $K\subseteq K_{n}$.

PatrickR points out the existence of a sequence of compact sets as you are describing is known as hemicompactness. Note that this is generally a bit weaker than being exhaustible by compact sets, though in the case of first countable spaces they are equivalent.

Finally, it is probably worth pointing out that in the case of first countable Hausdorff spaces (including e.g. metrizable spaces), hemicompactness implies local compactness, and so is equivalent to local compactness plus $\sigma$-compactness.

M W
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    The space $X=[0,\omega_1)$ of countable ordinals with the order topology is the union of an uncountable chain of compact sets $K_\alpha=[0,\alpha]$ and every compact subset is contained in one of them. – bof Jan 11 '24 at 03:48
  • Thanks for the answer! This is exactly what i needed. Do you know how to construct such $(K_n)_{n\in\mathbb{N}}$ in the case that $X$ is a Hausdorff locally compact $\sigma$-compact space? Or where could I find that construction? Thanks! – Susana Santoyo Jan 11 '24 at 03:59
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    @SusanaSantoyo If $X$ is $\sigma$-compact and every point has a compact neighborhood, then $X$ is the union of a sequence of compact sets $K_n$ such that $K_n\subseteq\operatorname{int}(K_{n+1})$. – user14111 Jan 11 '24 at 04:16
  • @SusanaSantoyo if you follow the link to pi-base above it gives a further link to this answer: https://math.stackexchange.com/questions/2078142/a-question-about-local-compactness-and-sigma-compactness/4568032#4568032 (bof's answer there, though not the one linked, gives the construction you are looking for in the locally compact Hausdorff case). – M W Jan 11 '24 at 04:50
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    @SusanaSantoyo The existence of such a sequence of $K_n$ is exactly the hemicompact property. See pi-base for relation with other properties. Note that hemicompact does not mean that any sequence of $K_n$ will work though, as M W has already shown, even in $\mathbb R^n$. – PatrickR Jan 12 '24 at 08:55
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Counterexample: $X = K = [0,1]$, $K_n = \{0\} \cup [1/n, 1]$.

Robert Israel
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    I was just writing $K_n = [0,1-{1 \over n}] \cup {1}$ :-) – copper.hat Jan 11 '24 at 03:00
  • Ok, it isn't always true. Thanks. However, here we can still find a sequence with that property ($K_n=X$ for all $n\in\mathbb{N}$). Do you know if such a sequence always exist? – Susana Santoyo Jan 11 '24 at 03:52