# Which compact metrizable spaces have continuous choice functions for non-empty closed sets?

Let $$X$$ be a compact metrizable space and let $$\mathcal{K}_{ne}(X)$$ be the collection of non-empty closed subsets of $$X$$ with the Vietoris topology (i.e. the topology induced by the Hausdorff metric for any compatible metric on $$X$$).

Question: When does there exist a continuous function $$f: \mathcal{K}_{ne}(X) \rightarrow X$$ such that for every $$G \in \mathcal{K}_{ne}(X)$$, $$f(G) \in G$$?

This feels like it should have been studied before, but I am unable to find a reference.

Some easy observations:

• If $$X$$ has a continuous choice function for non-empty closed sets and $$Y$$ is a closed subspace of $$X$$, then $$Y$$ has a continuous choice function for non-empty closed sets.

• $$\inf : \mathcal{K}_{ne}([0,1])\rightarrow [0,1]$$ is a continuous choice function for non-empty closed subsets of $$[0,1]$$. So we also have this for any closed subspace of $$[0,1]$$, such as Cantor space and any countable compact metrizable space.

• The circle and the tripod (three copies of $$[0,1]$$ glued together at $$0$$) both do not have continuous choice functions for non-empty closed sets (in both spaces given a set with two points there is a continuous path that makes the points switch places while keeping them separate). So no spaces in which these embed do either.

• Any finite disjoint union of spaces with continuous choice functions for non-empty closed sets also has a continuous choice function for non-empty closed sets (having elements in a clopen subset is a clopen condition in $$\mathcal{K}_{ne}(X)$$, so we can patch together the choice functions by cases).

A reasonable conjecture is that any such space embeds into $$[0,1]$$, but I could also see something tricky like the pseudo-arc having a continuous choice function for non-empty closed sets.

• Nice question! Let $N$ be the one-point compactification of $\mathbb N$. Then the same kind of argument (by considering two-point subsets) also proves that $[0,1]\times N$ does not admit a continuous choice functions for non-empty closed sets. Mar 3 '20 at 22:00
• @AndréHenriques Very nice. A similar argument should work for solenoids. Mar 4 '20 at 5:01
• Indeed, a solenoid contains $[0,1]\times N$ as a subspace. Mar 4 '20 at 15:31
• See mathoverflow.net/questions/74614 for the special case of continuous choice from two-point sets, which already excludes lots of spaces. Mar 4 '20 at 17:18

It's an old (1981) theorem by Jan van Mill and Evert Wattel (see this paper) that a compact space has a continuous selection iff it is orderable. (So has a linear order whose order topology is the topology on $$X$$). $$F \to \min(F)$$ and $$F \to \max(F)$$ are then the two only continuous selection functions IIRC. Even a continuous selecting function for $$[X]^2$$, the subspace of doubletons, is enough to get orderabilility.
If a space $$X$$ admits three distinct points $$x_1,x_2,x_3\in X$$ such that $$X\setminus\{x_i\}$$ is connected for every $$i=1,2,3$$, then $$X$$ does not admit a continuous choice function from the set of two-point subsets of $$X$$ back to $$X$$.
Indeed, such a choice function could be used to define three continuous maps $$f_i:X\setminus\{x_i\}\to \{0,1\}$$, and at least one of $$f_1,f_2,f_3$$ would have to be non-constant. Contradiction.
• Do you know if this implies that any continuum with more than one point and a continuous choice function for non-empty closed sets must be homeomorphic to $[0,1]$? Mar 4 '20 at 16:50
• It seems that the answer is yes. A continuum cannot have precisely 1 non-cut point and a continuum with precisely 2 non-cut points is homeomorphic to $[0,1]$. Mar 4 '20 at 16:56