A subset of $X$ an indecomposable continuum $Y$ is called a composant transversal if $X$ has exactly one point from each composant of $Y$.

So a continuum has a composant transversal precisely when there exists a choice function on its set of composants.

Solecki proved that no indecomposable continuum has a Borel composant transversal.

Solecki, Sławomir, The space of composants of an indecomposable continuum, Adv. Math. 166, No. 2, 149-192 (2002). ZBL1014.54021.

But what if we ask something a little different:

Main Question. Let $X$ be a dense Polish subspace of an indecomposable continuum $Y$. Must some composant of $Y$ contain at least two points of $X$?

If "Borel" replaces "Polish", then I have an example showing the answer is no. But I suspect the answer is yes. For instance, it is not difficult to show that a dense $G_\delta$-subset of $C\times[0,1]$ must contain uncountably many points from each of uncountably many arcs $\{c\}\times[0,1]$.

Ideas: If the union of all composants touching $X$ is Polish, then I think Solecki's results show the answer is yes. But the most I can show is that $X$ is a countable union of $G_\delta$-sets. Another idea is to suppose $X$ is a Polish "partial transversal", and then show $X$ can be extended to a Borel composant transversal, reaching a contradiction.

On another note, by Solecki's result the existence of composant transversals seems to require something close to the full-blown axiom of choice, or at least AC$(\mathbb R)$.

Other questions.

  • Are composant transversals non-measurable?

  • Does something like ZF$+$AD$+$DC imply there is no composant transversal?

  • $\begingroup$ Each Polish space contains a dense zero-dimensional Polish subspace (this follows from the Lavretiev Theorem). For this dense zero-dimensional subspace the composants are singletons. $\endgroup$ – Taras Banakh Aug 24 at 12:00

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