# Direct limit of compact topological spaces

Let $\{X_n\}_{n\in \mathbb{N}}$ be a direct system of compact topological spaces, meaning that we have morphisms $f_i\colon X_i \to X_{i+1}$ with the necessary compatibility conditions. Is there any chance that the direct limit $X$ is compact?

Any reference is welcome thanks

• Rarely? I mean, $\mathbb{N}$ is a filtered colimit of compact spaces, and so is $\mathbb{R}$. Can you be more specific about what you're looking for? Apr 3, 2015 at 19:31
• @QiaochuYuan Do you mean that if we take $X_n$ to be the set $\{1,\dots, n\}$ with the disccret topology and as maps the inclusions, then the limit is $\mathbb{N}$, which is not compact? Apr 3, 2015 at 19:41
• Yes, that's what I mean. Similarly, for $\mathbb{R}$ take the closed intervals $[-n, n]$. Apr 3, 2015 at 19:46
• So also T2 and connected is not enough; mine seems a hopeless question Apr 3, 2015 at 19:52

A $T_1$ colimit $X$ of a sequence of compact spaces $X_n$ is compact iff there is some $n$ such that the map $X_n\to X$ is surjective. This condition is obviously sufficient; suppose that it fails. Passing to a subsequence, we may assume that for each $n$, there is a point $x_n\in X$ that is in the image of $X_n$ but not the image of $X_{n-1}$. Choosing one such point for each $n$, we then get an infinite set $\{x_n\}\subseteq X$ such that for every subset $A\subseteq \{x_n\}$, the intersection of $A$ with the image of $X_n$ is finite for each $n$. In particular, since $X$ is $T_1$, this intersection is closed, so the preimage of $A$ in $X_n$ is closed. Since $X$ is the colimit of the $X_n$'s, this means that every subset of $\{x_n\}$ is closed in $X$. It follows that $X$ cannot be compact.

The $T_1$ hypothesis is necessary here; for instance, any Alexandrov space is the colimit of any sequence of subspaces that cover it. It is also very much necessary to have a sequential colimit rather than an arbitrary filtered colimit; for instance, any compact metric space is the colimit of its countable closed subspaces.