Suppose $X$ is a space and $A_1\subseteq A_2\subseteq A_3\subseteq ...\subset X$ is a sequence of subspaces each of which is closed in $X$ and such that $X\cong \varinjlim_{n}A_n$ (i.e. $U$ is open in $X$ if and only if $U\cap A_n$ is open in $A_n$ for each $n$). This topology on $X$ has many names (direct limit, inductive limit, weak topology, maybe more) but I can't seem to find much dealing with separation properties in this general setting. Specifically, I am asking:

If $A_n$ is Hausdorff for each $n$, then must $X$ also be Hausdorff?


1 Answer 1


The answer is no. H. Herrlich showed, in 1969, that even if you consider each $A_n$ a completely regular space, the direct limit may fail to be Hausdorff. However if all $A_n$ are T$_4$ - spaces then $X$ is a T$_4$ - space (it's not hard to prove this).

A comment about the definition of direct limit. Usually, in category theory, we call direct limit a colimit of a directed family of objects. Using this terminology it's well known that the category of Hausdorff spaces isn't closed under direct limits. You can find some examples in Dugundji's 'Topology' (a shame it's out of print). The definition you are using is very particular, so Herrlich's example is special.

In this paper D. Hajek and G. Strecker exhibit sufficient conditions for the Hausdorff property to be preserved under direct limits.

  • $\begingroup$ Very nice. I was hoping that when the maps in the directed limit were closed embeddings nice things would happen but these papers treat this perfectly. Thanks! $\endgroup$
    – J.K.T.
    Nov 3, 2010 at 5:15
  • $\begingroup$ The link to Herrlich's paper is broken, here is a working replacement. $\endgroup$
    – J. C.
    Jun 15, 2023 at 17:56

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