Let $X $ be a compact topological space (not necessarilly Hausdorff). I am looking for a charactrization for the following property:

Property: If $C $ is a closed subset of $X $, then there are pairwise disjoint closed subsets $C_1$,...,$C_n $ of $X $ such that $C=C_1\cup\dots\cup C_n $, and each $C_i $ has the property that if $A $ is a clopen subest of $C_i $, then there exists a clopen subset $B $ of $X $ with $A=C_i\cap B$?

Any comment is very welcome.

  • 4
    $\begingroup$ The spaces that every closed subset has finitely many connected components are example of such spaces. $\endgroup$
    – E.R
    Commented Jan 16, 2019 at 17:21
  • 2
    $\begingroup$ Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness. $\endgroup$ Commented Jan 16, 2019 at 19:02
  • $\begingroup$ By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”? $\endgroup$ Commented Jan 16, 2019 at 22:29
  • $\begingroup$ @Luke, yes in the subspace topology. $\endgroup$
    – Andy.G
    Commented Jan 17, 2019 at 2:35
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    $\begingroup$ Neither direction is obvious; I'm writing an answer elaborating now. $\endgroup$ Commented Jan 17, 2019 at 2:51

1 Answer 1


I don't see any reason to believe there is a simpler necessary and sufficient condition in general. When $X$ is Hausdorff there is one, though: your condition is equivalent to $X$ being totally disconnected.

To prove this, first suppose $X$ is compact Hausdorff and totally disconnected. Then clopen sets separate points of $X$ (see Any two points in a Stone space can be disconnected by clopen sets). It follows by compactness that clopen sets also separate closed sets (the proof is the same as the proof that a compact Hausdorff space is normal, just using clopen sets instead of open sets everywhere). So in particular, if $C\subseteq X$ is closed and $A$ is clopen in $C$, then $A$ and $C\setminus A$ can be separated by clopen subsets of $X$ (since they are disjoint and closed in $X$). This means that your condition holds, with $n=1$.

Conversely, suppose $X$ is compact Hausdorff and not totally disconnected. Your property is inherited by closed subspaces, so we may replace $X$ with one of its nontrivial connected components. So, we assume $X$ is compact Hausdorff and connected and has more than one point, and must show it does not satisfy your condition.

To prove this, pick an infinite discrete subset $D\subset X$ (see Every infinite Hausdorff space has an infinite discrete subspace) and consider $C=\overline{D}$. Suppose a decomposition $C=C_1\cup\dots\cup C_n$ with your property existed. Note that each point of $D$ is isolated in $C$, and some $C_i$ must contain infinitely many points of $D$. In particular, some $C_i$ must be disconnected, so there is a nontrivial clopen subset $A\subset C_i$. But $X$ is connected, so it has no nontrivial clopen subsets, and so $A$ cannot be $B\cap C_i$ for any clopen $B\subseteq X$.

Of course, total disconnectedness is not necessary in the non-Hausdorff case (I don't know whether it is sufficient). Besides trivial examples like the indiscrete topology, there is also the cofinite topology on any set. More generally, as Es.Ro commented, a sufficient condition is that every closed subset of $X$ has only finitely many connected components, since then you can take the connected components as the $C_i$.

  • $\begingroup$ Thanks for your answer. $\endgroup$
    – Andy.G
    Commented Jan 17, 2019 at 3:00
  • $\begingroup$ Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property? $\endgroup$
    – Andy.G
    Commented Jan 17, 2019 at 14:55

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