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Let $X$ be a compact metric space and $P$ be a path component of $X$. Since we are not assuming $X$ is locally path connected, $P$ must need not be open nor closed. Certainly, $P$ must be separable but how bad can $P$ be? Is every separable path-connected metric space, homeomorphic to the path component of some compact metric space?

I'm curious about this question because subspaces like $w(Y)=\{x\in X\mid \text{$X$ is not semilocally simply connected at $x$}\}$ are important homotopy-invariants in the study of Peano continua. If $Y$ is compact and locally path-connected, $X=w(Y)$ may be any compact metric space. Iteration of $w(-)$ becomes relevant. By this, I mean take a path component $P$ of $X$ and then take $w(P)$. This is why I want to know how bad $P$ can be. However, I don't know much about metric compactifications that would be relevant.

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    $\begingroup$ When you say "Is every separable..." you probably mean "Is every separable...homeomorphic to a path component..." $\endgroup$
    – Moishe Kohan
    Commented May 29, 2021 at 19:07
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    $\begingroup$ Surely you need to assume that your space is path connected in order to hope for it to be a path component. $\endgroup$
    – Wojowu
    Commented May 29, 2021 at 19:11
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    $\begingroup$ does every separable metric space even have a metrizable compactification? I am not sure if this is useful, but: if your space $X$ is locally compact and has a metrizable compactification $Y$, then $X$ is open in $Y$, and the closure in $Y\times [-1,1]$ of the graph of $\sin (\frac{1}{d(x, Y\backslash X)})$, $x\in X$ is what you need (it seems). $\endgroup$
    – erz
    Commented May 29, 2021 at 22:50
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    $\begingroup$ @erz I like this construction. In fact, you can use the one-point compactification if $X$ is locally compact and separable. This does seem to give a partial answer: every path-connected, locally compact, separable metric space is the path component of some compact metric space. $\endgroup$
    – Jeremy Brazas
    Commented May 31, 2021 at 13:19
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    $\begingroup$ A path component of a compact metric space is in general an analytic set (=continuous image of a Polish space), but it need not to be a Borel set (see a paper by Becker, 1998: The number of path-components of a compact subset of $\mathbb R^n$, Corollary 4.2). For path components of compact subsets of $\mathbb R^2$, something more can be said. $\endgroup$
    – Benjamin Vejnar
    Commented Jun 2, 2021 at 11:53

1 Answer 1

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Not every path connected separable metric space is homeomorphic to a path component of a compact metric space. The following cardinality arguments can be used:

Fact 1. There is up to homeomorphism more than $|\mathbb R|$-many path-connected separable metric spaces: Just consider all the subspaces of the form $X_A=(\mathbb R\times (0,1))\cup (A\times\{0\})$ for $A\subseteq \mathbb R$. Each of them is path-connected and every space $X_A$ is homeomorphic to at most $|\mathbb R|$-many spaces $X_B$.

Fact 2. There is up to homeomorphism only $|\mathbb R|$-many analytic subsets of compact metric spaces.

Fact 3. Every path-connected component of a compact metric space is analytic.

Hence there is at least one path connected space of the form $X_A$ which is not homeomorphic to a path component of a compact metric space.

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  • $\begingroup$ This is great ! $\endgroup$
    – Joel David Hamkins
    Commented Jun 2, 2021 at 14:33
  • $\begingroup$ It could be possibly true that every path-connected separable analytic metric space is homeomorphic to a path component of a compact metric space. $\endgroup$
    – Benjamin Vejnar
    Commented Jun 3, 2021 at 14:40
  • $\begingroup$ So any path-connected, separable, but non-analytic metric space would do. Really nice! If you find an argument with for the possibility mentioned in your comment, please let me know. $\endgroup$
    – Jeremy Brazas
    Commented Jun 3, 2021 at 20:06

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