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?
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?
Best Answer
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 $T_1$.
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;
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. $\;\square$