I apologize in advance if this question is thought to be too soft or otherwise inappropriate for mathoverflow.net. Let M be the infinite set of all homeomorphism types of finite dimensional topological manifolds which are compact, connected and metrizable. Many years ago, the question of whether M was countable, was an open question. Eventually the answer was proved to be "Yes". Are there any examples of well known infinite sets (in this or that branch of mathematics) for which the question of whether the set is countable or uncountable is still considered an interesting open problem?

$\begingroup$ This is not an answer as it is not about a particular set, but I’m reminded of the (still wide open) conjecture due to Zilber, which states that all sets firstorder definable in the complex exponential field are either countable or have a countable complement. $\endgroup$ – Emil Jeřábek supports Monica Dec 7 '14 at 21:08

4$\begingroup$ It seems misleading to call that a question about infinite sets. Of course it is a question about manifolds. $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 22:37

3$\begingroup$ How about the largest infinite set of cardinality less than the continuum? $\endgroup$ – Gerry Myerson Dec 8 '14 at 0:24

$\begingroup$ That's clever and a hard act to follow. $\endgroup$ – Garabed Gulbenkian Dec 11 '14 at 20:07
Here is a question of classical function theory of this sort. Let $f$ be an entire function of one variable. Consider all factorizations $f=g\circ h$ where $g$ and $h$ are entire. Two factorizations $f=g_1\circ h_1$ and $f=g_2\circ h_2$ are called equivalent if there is a polynomial $L$ of degree $1$ such that $g_2=g_1\circ L$ and $h_2=L^{1}\circ h_1$.
Can an entire function have uncountably many nonequivalent factorizations?
An example which has infinitely many is $e^z$ with $g_n(z)=z^n$ and $h(z)=e^{z/n}$. This was unsolved when I was a student in 1970th, and I suppose still is.
EDIT. Another question of this sort from the theory of entire functions was this (Erdos). Consider a family $A$ of entire functions with the property that for every $z\in C$, the set $\{ f(z):f\in A\}$ is at most countable. Does it follow that $A$ is at most countable? But this question has been "solved": it turns out that the answer depends on the Continuum Hypothesis:)

$\begingroup$ But is there a specific entire function which is known to have infinitely many nonequivalent factorizations, but for which it is not known whether that infinity is countable? $\endgroup$ – Gerry Myerson Dec 8 '14 at 0:21


$\begingroup$ Many thanks for your responses. I had never heard of any of these examples before. $\endgroup$ – Garabed Gulbenkian Dec 11 '14 at 19:44
Until very recently, it was unknown whether or not there are uncountably many topological isomorphism classes of compactly generated simple nondiscrete locally compact groups. (There are, thanks to a construction of Simon Smith that appeared on the arXiv earlier this year.) I think it is still unknown whether there are uncountably many local isomorphism classes of them.
Also, I am not sure about this, but I think it is still unknown whether or not there are uncountably many commensurability classes of hereditarily just infinite pro$p$ groups.