Compact sets of direct limit topology

Question

We consider a direct limit of a tower $X_1\subset\cdots\subset X_n$ of spaces, where each $X_n$ is a subspace of $X_{n+1}$. The direct limit is $X_\infty:=\bigcup_n X_n$ endowed with the topology $\mathcal{T}_{\infty}$ defined as follows: $U\subset X_\infty$ is open if and only if $\forall n\in \Bbb{N},\quad U\cap X_n$ is open on $\mathcal{T_n}.$

We assume that $\mathcal{T}_n$ is the subspace topology on $X_n$ by $\mathcal{T_{n+1}}.$

If I denote $\mathcal{K}_n$ the family of compact sets of $X_n$ then the union of such sets are included in $K_\infty$ (which is the family of compact sets of $X_{\infty}$

If $K$ is compact of $X_n$ then it's a compact of $X_\infty.$ Do we have equality?

In other words the collection of all compact subsets of $X_{\infty}$ is the union of the corresponding collections for each $X_n.$

EDIT Thanks to Eric Wofsey the answer is no if $X_{\infty}$ is not Hausdorff. What if it's Hausdorff ? (we say $T_0$?)

Answer 1

No, not necessarily. For instance, suppose each $X_n$ has the indiscrete topology. Then $X_\infty$ will also have the indiscrete topology, and every subset of $X_{\infty}$ is compact. So as long as $X_\infty$ is not equal to $X_n$ for any $n$, then $K=X_\infty$ is a compact subset that is not contained in any $X_n$.

Answer 2

It's quite easily true if every $X_n$ is open in $X_{n+1}$: then the $X_n$ will form an open cover of $X_\infty$, so a member $K$ of $K_\infty$ is covered by finitely hence by one $X_n$, and so $K \subseteq X_n$ for some $n$.