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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.

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  • $\begingroup$ 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. $\endgroup$ – André Henriques Mar 3 at 22:00
  • $\begingroup$ @AndréHenriques Very nice. A similar argument should work for solenoids. $\endgroup$ – James Hanson Mar 4 at 5:01
  • $\begingroup$ Indeed, a solenoid contains $[0,1]\times N$ as a subspace. $\endgroup$ – André Henriques Mar 4 at 15:31
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    $\begingroup$ See mathoverflow.net/questions/74614 for the special case of continuous choice from two-point sets, which already excludes lots of spaces. $\endgroup$ – Andreas Blass Mar 4 at 17:18
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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.

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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.

This argument excludes the pseudo-arc.

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  • $\begingroup$ 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 $[0,1]$? $\endgroup$ – James Hanson Mar 4 at 16:50
  • $\begingroup$ 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 $[0,1]$. $\endgroup$ – James Hanson Mar 4 at 16:56
  • $\begingroup$ @JamesHanson a metrisable continuum, yes, let's not restrict ourselves... $\endgroup$ – Henno Brandsma Mar 6 at 23:33

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