Without Hausdorff, what implications can we prove about $k_\omega$ related to other covering properties?

Question

In e.g. A SURVEY OF $k_\omega$-SPACES a space is said to be $k_\omega$ if it's the union of compact Hausdorff $K_n$, $n<\omega$, with a set being closed if and only if its intersection with each $K_n$ is closed.

It's asserted there that $k_\omega$ is implied by $K_n\subseteq int(K_{n+1})$, i.e. exhaustible by (Hausdorff) compacts.

I want to consider the case where the $K_n$ need not be Hausdorff. In this case, we have exhaustible by compacts $\Rightarrow$ hemicompact $\Rightarrow$ $\sigma$-compact (with no arrows reversing). Where does this non-$T_2$ $k_\omega$ live?

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Exhaustible by compacts implies this $k_\omega$. Let $K_n\subseteq int(K_{n+1})$. $k_\omega$ is equivalent to a set having open intersection with each $K_n$ with respect to the subspace topology implies the set is open. So let $U$ have open intersection with each $K_n$ and let $x\in U$. Note $x\in K_n$ for some $n<\omega$. Then let $V$ be an open set such that $V\cap K_{n+1}=U\cap K_{n+1}$. $x\in V\cap int(K_{n+1})\subseteq U$, proving $U$ is open.

Obviously, $k_\omega$ still implies $\sigma$-compact. Where does hemicompact fit in without something like Hausdorff or locally compact?

Answer 1

Edit: My original post contained a faulty counterexample with a rather obvious flaw. Since the rest of the post contained some useful (but well-known) material, I will leave this up as an extended comment.

So you want to consider a space $X$ which is the colimit of an expanding sequence $$X_1\subseteq X_2\subseteq\dots\subseteq X_n\subseteq\dots$$ of its compact (not necessarily Hausdorff) subspaces? Such an $X$ is Lindelöf, and is $T_1$ when each $X_n$ is $T_1$.

In case $X$ is $T_1$, it is hemicompact.

Proof: Suppose that $K\subseteq X$ is a compact subset not contained in any $X_n$. We can assume without loss of generality that the $X_n$ are all distinct and that $K\cap X_1\neq\emptyset$. For each $n\geq0$ choose $x_n\in K\cap(X_n\setminus X_{n-1})$, where we understand $X_{0}=\emptyset$. Write $D=\{x_n\mid n\in\mathbb{N}\}\subseteq X$. If $C\subseteq D$ is any subset, then $C$ is closed in $D$, since for each $n$ we have that $C\cap X_n$ is finite and hence closed in $X_n$. It follows that $D$ is infinite, discrete and closed in $X$, and this contradicts the compactness of $K$. $\;\blacksquare$

On the other hand, with not separation assumptions we have;

If $X_n$ is open in $X_{n+1}$ for each $n\in\mathbb{N}$, then $X$ is hemicompact.

Proof: The family $\{X_n\}_{n\in\mathbb{N}}$ is a an open cover of $X$. $\;\blacksquare$

Example 0: A hemicompact example which is not exhaustible by compacts.

For $n\in\mathbb{N}$ let $X_n=\bigvee^n_{i=1}S^1$ and equip $X=\bigcup^\infty_{i=1}X_n$ with the colimit topology. As a countable CW complex, $X$ is a Hausdorff $k_\omega$-space. If $X$ had a sequence of exhaustion $K_1\subseteq int(K_2)\subseteq K_2\dots$, then each point of $X$ would be contained in the interior of one of the compact $K_n$'s. Since each compact subset of a CW complex is metrisable, each point of $X$ would therefore have a metrisable neighourhood. But a locally metrisable paracompact space is metrisable, and $X$ is not even first-countable at the wedge point. $\;\square$

Answer 2

Not an answer, but some obervations. Hereafter $k_\omega$ will mean the version without Hausdorff.

A hemicompact space need not be $k_\omega$, even in the Hausdorff case. The survey article gives as an example on p. 114 a certain subspace of the Cech-Stone compactification $\beta\mathbb{N}$.

Another example is the Arens-Fort space. As shows here this space $X$ is hemicompact. But it is not $k_\omega$. Indeed, it is anticompact and $T_1$, so every compact subspace is finite with the discrete topology. So every subset $A\subseteq X$ meets every compact subspace $K\subseteq X$ in a closed subset of $K$, but not every subset of $X$ is closed since the space is not discrete. This shows that $X$ does not admit any $k_\omega$ decompsition $(X_n)$.

So the other question is: does $k_\omega$ imply hemicompact?

The tricky thing seems that given a $k_\omega$ decomposition $(X_n)$, it is not necessarily the case that the space will be hemicompact with respect to that decomposition, but it could still be hemicompact with respect to a different decomposition.

Example 1 from Tyrone (now edited out, but available in the edit history) illustrates this. Taking $X=\mathbb{N}=\{1,2,...\}$ with the indiscrete topology, the sets $X_n=[1,n]$ (in the usual order on $X$) form a $k_\omega$ decomposition but not every compact subset ($X$ in particular) is contained in one of the $X_n$. But $X$ is still hemicompact because it is compact.

Another example is $X=\mathbb{Z}$ with the right order topology. Open sets are of the form $[n,\infty)$ and compact sets are the ones with a minimum element. The sets $X_n=[-n,n]$ form a $k_\omega$ decomposition (easy to check the weak topology condition), but the compact set $[0,\infty)$ is not contained in any of the $X_n$. Note that in this case the closure of any of the $X_n$ is not even compact. But $X$ is still hemicompact by taking $X_n=[-n,\infty)$ for example.

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(Added 12/11/2022): The paper https://arxiv.org/abs/1602.04857 by T. Banakh uses the notions of $k_\omega$ and hemicompact spaces without assuming Hausdorff.

On the bottom of p. 8 it states:

"Each $k_\omega$-space is hemicompact."

Also on p. 16 before Lemma 2.5.1 it restates the same thing:

"It is known that each $k_\omega$-sequence $(K_n)_{n\in\omega}$ in a topological space X swallows compact subsets of the space in the sense that each compact subset $K\subset X$ is contained in some set $K_n$."

But this last sentence cannot be true in general without Hausdorff, as my two examples above show. Could be worth asking the author about it.

Answer 3

A partial result adapting 4.3 of this paper assuming the space is first-countable and $T_1$. Let $K_n$ for $n<\omega$ witness some $k_\omega$ decomposition with $K_n\subseteq K_{n+1}$. Suppose, aiming for a contradiction, some $p\in X$ has no compact neighborhood; in particular, no $K_n$ is a compact neighborhood of $p$. Without loss of generality, assume $p\in K_0$. Then if $\{U_n:n<\omega\}$ is the guaranteed (by first-countability) countable base at $p$, $U_n\not\subseteq K_n$; pick $p_n\in U_n\setminus K_n$. Then $\{p_n:n<\omega\}$ is closed as it has finite (a subset of $\{p_m:m<n\}$, and therefore closed by $T_1$) intersection with each $K_n$. But $p\not\in\{p_n:n<\omega\}$ is a limit point as $p_n\in U_n$ for all $n<\omega$, a contradiction.

So every first-countable $T_1$ $k_\omega$ space is weakly locally compact. Since weakly locally compact and $\sigma$-compact spaces are hemicompact, and first-countable hemicompact spaces are all weakly locally compact, $k_\omega$ and hemicompact are equivalent for first-countable $T_1$ spaces.

Answer 4

I realize this is a little old but I see the question is not resolved yet. Here is the answer.

Definition. $X$ is generated by compact sets ($CG1$) if for each subset $A\subseteq X$, $A$ is closed in $X$ if and only if $A\cap K$ is closed in $K$ for each compact $K$.

Proposition. Let $X$ be a topological space. Then $X$ is $k_\omega$ if and only if $X$ is $CG1$ and hemicompact.

Proof.

If $X$ is hemicompact and $CG1$, with hemicompactness witnessed by a sequence $\{K_n\}$, then suppose $A\subseteq X$, with $A\cap K_n$ closed in $K_n$ for each $n$. Then for any compact $K$, choose $n$ with $K\subseteq K_n$. Since $A\cap K_n$ is closed in $K_n$, $A\cap K= (A\cap K_n)\cap K$ is closed in $K$. Thus by $CG1$, $A$ is closed in $X$, and so the sequence $K_n$ witnesses that $X$ is $k_\omega$.

Conversely, we first observe that the argument for 3) on page 113 of the linked survey would work fine if we restricted ourselves to $T_1$ spaces. For the general case, we modify it as follows.

Let $X$ be $k_\omega$, witnessed by $\{K_n\}$. For each $n$, let $L_n=\{x\in X\mid \overline{\{x\}}\cap K_n\neq \emptyset\}$. We first observe that $L_n\supseteq K_n$, and $\mathcal U$ is an open cover of $K_n$ if and only if it is an open cover of $L_n$. Indeed, if $\mathcal U$ covers $K_n$, and $x\in L_n$, then there is some $y\in\overline{\{x\}}\cap K_n$, and some $U\in \mathcal U$ with $y\in U$. Then since $y\in \overline{\{x\}}$ and $U$ is a neighborhood of $y$, $U$ meets $\{x\}$, i.e., $x\in U$. The other direction is trivial.

As a result, compactness of $K_n$ immediately implies compactness of $L_n$. Note that $L_n$ is certainly an increasing sequence of compact sets. Now suppose $K\subseteq X$ is compact. We wish to show $K\subseteq L_n$ for some $n$, so supposing otherwise, choose $x_n\in K\backslash L_n$ for each $n$. Then by definition, we have $\overline{\{x_n\}}\cap K_n=\emptyset$.

Now define $U_n=X\backslash\bigcup_{k\geq n} \overline{\{x_k\}}$. Then for each $n$ and $m$, $U_n\cap K_m$ is open in $K_m$, since only finitely many of the closed sets $\overline{\{x_k\}}$ intersect $K_m$. Since this holds for all $m$, by the $k_\omega$ condition, $U_n$ is open. But then $\{U_n\}$ is an open cover of $K$ with no finite subcover, contradicting compactness.

Thus $X$ is hemicompact via the sequence $L_n$, and of course $k_\omega$ trivially implies $CG_1$.

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Remarks.

Note that in contrast to the non-$T_1$ examples in PatrickR's answer, under $T_1$, the $k_\omega$ and hemicompact properties use the same sequence, since under $T_1$ we have $L_n=K_n$ in the above proof.

In general, we have the implications

$$\text{exhaustible-by-compacts}\implies k_\omega\implies \text{hemicompact}\implies \sigma\text{-compact},$$

all strict (even under separation as strong as $T_6$), with the converse of the first refuted by the Arens space (or any of the Fréchet-Urysohn counterexamples discussed here), the converse of the second refuted by the Arens Fort space per PatrickR's answer, and the converse of the third refuted by many examples, including $\mathbb Q$.