Is there a closed totally disconnected space meeting every component?

Question

Suppose $X$ is compact Hausdorff and has infinitely many components. It seems there should be a way to select at least one point from each component so the resulting set is totally disconnected and closed. But it seems harder than I thought.

Brute force approach: Let $C$ denote the family of components of $X$. For each $c \in C$ choose a point $x(c) \in c$. Then the set $D = \{x(c): c \in C \}$ has only one point from each component. But it might not be closed. So redefine $D$ as the closure of $\{x(c): c \in C \}$. Then $D$ is closed but might not be totally disconnected.

Any tips?

Answer 1

Consider $$X=A\cup B$$ with $$A=\left\lbrace\left(\frac1n,\sin n\right),n\in\Bbb N\right\rbrace\qquad \text{and} \qquad B=\{0\}\times\left[-1,1\right]$$

This is compact Hausdorff, but any closed subset meeting once each component would have to contain the whole of $A$ and therefore $B$ as well.

Answer 2

To complement Arnaud Mortier's counterexample, it is possible to do so at least when $X$ is second countable or equivalently metrizable and the decomposition of $X$ to its components is a continuous (or equivalently lower semicontinuous) decomposition.

A decomposition $C$ of $X$ is upper semicontinuous if for every open set $U ⊆ X$ the set $⋃\{c ∈ C: c ⊆ U\}$ is open as well. And it is lower semicontinous if for every open set $U ⊆ X$ the set $⋃\{c ∈ C: c ∩ U ≠ ∅\}$ is open. Finally, it is continuous is it is both upper and lower semicontinuous.

Let $q: X \to C$ be the map such that $x ∈ q(x)$ for every $x ∈ X$. Let us endow $C$ with the quotient topology, so $q$ becomes a quotient map. It turns out that $C$ is upper semicontinuous iff $q$ is closed, and it is lower semicontinous iff $q$ is open.

In your particular case when $C$ is the decomposition of a Hausdorff compact space into its components, $C$ is not only hereditarily disconnected, but even totally separated, and so zero-dimensional. This is because components and quasi-components of $X$ are the same. In particular, $C$ is also a Hausdorff compact space, and $q$ is closed.

If additionally, $q$ is open or equivalently $C$ is lower semicontinuous, we may use Michael's zero-dimensional selection theorem to obtain a continuous selection $s:C \to X$. Being a selection means that $s(c) ∈ c$ for every $c ∈ C$. Therefore the image of $s$ is a closed subset of $X$ meeting every component at one point. It is also homeomorphic to $C$.

We shall conclude with the formulation of Michael's zero-dimensional selection theorem: Let $X$ be a paracompact zero-dimensional space, and let $Y$ be a completely metrizable space. Every lower semicontinuous map $F:X \to \mathcal{P}(Y)$ such that every $F(x)$ is closed and nonempty admits a continuous selection.

I realize that I didn't define semicontinous functions, but here $q$ is lower/upper semicontinuous iff $C$ is lower/upper semicontinuous.

[E. Michael, Selected Selection Theorems, The American Mathematical Monthly, Vol. 63, No. 4. (Apr., 1956), pp. 233-238]