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We say that a topological space $(X,\tau)$ is flexible, if for every closed discrete subset $D\subseteq X$ and every map $f: D\to X$ there is a continous map $f^X:X\to X$ such that $f^X|_D = f$.

$\mathbb{R}$ with the Euclidean topology is flexible.

Is every homogeneous $T_2$-space flexible?

Note. I called spaces with the above property "flexible" because they are in some sense the opposite of (strongly) rigid spaces; flexible spaces admit a lot of continuous self-maps, and strongly rigid spaces only very few.

Dominic van der Zypen
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1 Answers1

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I like this question a lot.

The answer is no. Let $X$ consist of two disjoint copies of the real line. This is $T_2$ and homogeneous, in the sense that for any two points, there is a homeomorphism taking the first to the second — all points look alike.

But it is not flexible, since if $D$ consists of two points $a$ and $b$, both in the same copy, and we let $f(a)$ be in one copy and $f(b)$ in the other, we cannot extend this to a continuous map on the whole space, since the image of the copy of $\mathbb{R}$ from which $a$ and $b$ came must be connected, but it cannot be.

(This answer suggests that one should consider a stronger notion of homogeneity, for example, insisting not just that the group of self-homeomorphisms acts transitively, but that every finite permutation extends to a self-homeomorphism. For example: if a space is flexible with respect to finite sets $D$, then is it flexible with respect to all closed discrete sets?)

Joel David Hamkins
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  • Thanks Joel for this beautiful example and for suggesting having a look at a stronger notion of homgeneity! – Dominic van der Zypen Jul 03 '18 at 12:15
  • Note that $\mathbb{R}$ is not homogeneous with respect to that stronger notion, since you can move one point from one side of two points to between them, but this will not extend to a homeomorphism. See also your earlier question: https://mathoverflow.net/a/245319/1946. – Joel David Hamkins Jul 03 '18 at 12:24