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(I am most interested in the case $X=\mathbb R^2$, but of course one could ask the same question for manifolds, or metric spaces in general.)

Let $\text{Com}(\mathbb R^2)$ denote the space of nonempty compact subsets of the plane, equipped with the Hausdorff metric. Let $S_\bullet:[0,1]\to\text{Com}(\mathbb R^2)$ be a continuous path, and let $p\in S_0$. Must there exist a path $\gamma:[0,1]\to\mathbb R^2$ such that $\gamma(0)=p$, and $\gamma(t)\in S_t$ for all $t\in[0,1]$?

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    $\begingroup$ I believe if you consider a "snake" going along closed topologist's sine curve for $t\in[0,1)$ and ending with the interval $\{0\}\times[-1,1]$ will give a continuous path of compacts with no path interpolating it. $\endgroup$
    – Wojowu
    Commented Oct 21, 2020 at 9:33
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    $\begingroup$ By the Kuratowski and Ryll-Nardzewski selector theorem you should be able to get a Borel path $\gamma\colon[0,1]\to X$ with $\gamma(t)\in S_t$ for all $t$ as long as $X$ is Polish. $\endgroup$
    – Alessandro Codenotti
    Commented Oct 21, 2020 at 9:35

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$\DeclareMathOperator{\R}{\mathbf{R}}\DeclareMathOperator{\Z}{\mathbf{Z}}$The answer is no, even in the circle (and hence in the plane).

As coordinates, write the circle as the 1-point compactification $\bar{\R}$ of $\R$.

For $t\in\mathopen]0,1]$, write $$X_t=\{\infty\}\cup\big(t\Z+\sin(1/t)\big).$$ For $t\to 0$, this tends to $X_0=\bar{\R}$, hence defines a continuous path on $[0,1]$. Any continuous lift, for $t>0$ has to have the form $x(t)=tn+\sin(1/t)$ for some fixed $n$. This does not converge when $t\to 0$ (it accumulates to all of $[-1,1]$). So there is no continuous lift.

(Note that the answer is clearly positive when $X=\mathbf{R}$, as $x\mapsto \max(x)$ is then a lift (not only for paths).)

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  • $\begingroup$ Yes I see, thanks. $\endgroup$
    – user2520938
    Commented Oct 21, 2020 at 10:51
  • $\begingroup$ I think there may be a problem with the limit set $X_0$ (or possibly I am misunderstanding your definition of $X_t$). One can choose a sequence $\{t_n\} \subseteq (0,1)$ such that $t_n\to 0$ and $\sin(1/t_n)=0$ for all $n$. For this sequence we have $X_{t_n} \to \{\infty\}\cup\{0\}$. $\endgroup$
    – Logan Fox
    Commented Oct 22, 2020 at 22:33
  • $\begingroup$ @LoganFox thanks for pointing out, I messed up $t\to 0$ with $t\to\infty$... I edited: the group term is $t\mathbf{Z}$ (and not $t^{-1}\mathbf{Z}$), so it's more and more dense when $t\to 0$. $\endgroup$
    – YCor
    Commented Oct 23, 2020 at 6:34

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