Polish transversals

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?

• 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. – Taras Banakh Aug 24 at 12:00