# Compact spaces in which any closed set can be partitioned into finitely many closed sets whose clopen subsets extend to the whole space

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.

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

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 $$C_i$$.

• 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? Commented Jan 17, 2019 at 14:55