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

  • 8
    $\begingroup$ 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? $\endgroup$ Apr 3, 2015 at 19:31
  • $\begingroup$ @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? $\endgroup$
    – Giulio
    Apr 3, 2015 at 19:41
  • $\begingroup$ Yes, that's what I mean. Similarly, for $\mathbb{R}$ take the closed intervals $[-n, n]$. $\endgroup$ Apr 3, 2015 at 19:46
  • 1
    $\begingroup$ So also T2 and connected is not enough; mine seems a hopeless question $\endgroup$
    – Giulio
    Apr 3, 2015 at 19:52

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.