Given a continuum $X$ (compact metrizable connected $X$) let $K(X)$ denote the hyperspace of nonempty compact subspaces of $X$ with the Vietoris topology and let $C(X)$ denote the (closed) subspace of $K(X)$ consisting of connected sets.

It is well known (and has been discussed on MO before) that if there is a continuous choice function $K(X)\to X$ then $X\simeq[0,1]$, indeed any other continua doesn’t even have a continuous choice function for the subspace of $K(X)$ of unordered pairs.

The situation is different when looking at $C(X)$: for example the “star” $X$ obtained by joining $n$ arcs at their endpoint has a continuous selection for $C(X)$ by looking at the tree partial order induced by choosing the common point as root and then mapping $C\in C(X)$ to its $\min$ with respect to this partial order. In a similar way we get a continuous selection for $C(X)$ whenever $X$ is a dendrite.

It is also easy to see that $S^1$ doesn’t have a continuous selection for $C(S^1)$ by using a combination of Borsuk-Ulam, $C(S^1)\simeq D^2$ and $\partial C(S^1)\cong F_1(S^1)$, where $D^2$ is the unit disk in $\Bbb R^2$ and $F_1(X)$ is the space of singletons of $X$.

This suggests the following: suppose $X$ is a continuum with a continuous selection $C(X)\to X$. Must $X$ be a subcontinuum of a dendrite? If not is there a characterization of the continua $X$ with a continuous selection $C(X)\to X$?

  • 2
    $\begingroup$ Presumably you exclude $\emptyset$ from $K(X)$. $\endgroup$ Sep 15 at 10:38
  • 3
    $\begingroup$ You can find more in the book by Illanes and Nadler: Hyperspaces, pages 363 and beyond. It is known that every selectible continuum is a dendroid, but there exist selectible dendroids which are not dendrites (and thus not a subcontinuum of dendrite, because a subcontinuum of a dendrite is a dendrite again). $\endgroup$ Sep 15 at 10:42
  • 1
    $\begingroup$ @BenjaminVejnar that should be an answer (there is no precise characterization of selectible continua according to the same book). (I also feel kind of dumb now, indeed my question was inspired by an exercise in an earlier chapter of the same book, and I completely missed that they have more to say on this topic later on) $\endgroup$ Sep 15 at 12:47
  • $\begingroup$ What's the definition of choice/selection function on $C(X)$? $\endgroup$
    – YCor
    Sep 15 at 13:07
  • 2
    $\begingroup$ @WillBrian Yes, that is example 75.9 in the book by Illanes and Nadler, the construction is originally due to Mackowiak. $\endgroup$ Sep 15 at 13:41

I will post a CW answer to my own question to remove it from the unanswered list since Benjamin doesn't seem interested in turning his comment into an answer, but this is all informations taken from the section of Hyperspaces that he suggested.

To begin with let's fix the terminology selectible to refer to a continuum $X$ for which there exist a continuous selection $C(X)\to X$.

The simple answer to the question at the end, namely "is there a characterization of the selectible continua?" is no.

What is known (according to the book by Illanes and Nadler, I'm not aware of any recent developement however) is more or less the following:

  • If $X$ is a selectible continuum, then $X$ is a dendroid.
  • Say that a selection $s\colon C(X)\to X$ is rigid if whenever $A,B\in C(X)$ are such that $s(B)\in A\subseteq B$, then $s(A)=s(B)$ (note that the definition of rigid selection in Hyperspaces has a typo, this is the correct definition as given in the original reference Rigid Selections and Smooth Dendroids by L.E.Ward). Then there is a characterization of the rigidly selectable continua: those are exactly the smooth dendroids by the paper of Ward mentioned above. The idea here is that if $X$ is a smooth dendroid then the "tree" partial order (with root a point witnessing smoothness) is closed in $X\times X$, so the map sending each subcontinuum to its minimum is a rigid selection. Conversely if $s\colon C(X)\to X$ is a rigid selection, then $\{\{s(A)\}\times A\mid A\in C(X)\}$ is a closed partial order on $X$ which agrees with the "tree" order with root $s(X)$, so that $s(X)$ is a point witnessing the smoothness of $X$.
  • Not all dendroids are selectible. Given two points $x,y$ in a dendroid let $[x,y]$ denote the unique arc joining them and let $(x,y)$ denote $[x,y]\setminus\{x,y\}$. Following definition 75.10 in Hyperspaces say that a dendroid $X$ is of type $N$ (a drawing justifies this name and clarifies the definition) if there are two points $p,q\in X$ and two sequences of arcs $[p_np_n'],[q_n,q_n']$ with points $p_n''\in(q_nq_n')$ and $q_n''\in(p_n,p_n')$ such that: $$\begin{align} [p,q] &=\lim [p_n,p_n']=\lim [q_nq_n'] \\ p &= \lim p_n = \lim p_n' = \lim p_n'' \\ q &= \lim q_n = \lim q_n' = \lim q_n''\end{align}.$$ Then any dendroid of type $N$ is nonselectible.
    A simple example of a dendroid of type $N$ is the doubly infinite broom: start with the segment between $(0,0)$ and $(1,0)$ in $\Bbb R^2$ and add segments between $(0,0)$ and $(1,1/n)$ and between $(1,0)$ and $(0,-1/n)$ for all $n\in\Bbb N^+$.
  • More surprisingly there are even contractible dendroids that are not selectible, this is example 75.9 in Hyperspaces and the construction is originally due to Maćkowiak.
  • selectibility is not preserved by neither monotone nor open maps. Nonselectibility isn't either, this is exercise 75.28 in Hyperspaces.
  • There are various open questions about selectibility apart from a classification of selectible continua, some of them are in chapter 76 of Hyperspaces.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.