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



Direct limit of compact topological spaces
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.


