Is it possible to prove in ZF that a non-trivial compact connected Hausdorff space is uncountable?

Let $X$ be a compact, connected Hausdorff space with at least two points.

In $\mathrm{ZF}+\mathrm{AC}_\omega(\mathbb R)$, any countable compact Hausdorff space is metrizable, and from this it can be shown that $X$ is uncountable.

In $\mathrm{ZF}$, however, that result does not hold. Does anyone know if it's still possible to prove that it is uncountable?

Choice is not needed.

EDIT 1: Thanks to @dfeuer for pointing out my original argument required Dependent Choice, and for helping me arrive at a choice-less proof.

EDITS 2&3: I discovered a new flaw in the `proof', namely that $$x^*$$ could be an element of $$C_n$$ (this should to be avoided to guarantee empty intersection of $$C_n$$'s). Will work to fix. Unfortunately in this "Leap Frog" argument it seems difficult to fix $$U$$ at stage $$n$$ without invoking DC. One solution would be to define set $$K$$ which cuts $$C_{n-1}$$ between $$x^*$$ and $$x^{**}$$. Then the $$U$$'s could be neighborhoods of $$K$$. I don't know if there's an explicit way to define a $$K$$.

I will say $$X$$ is countable if there is an injection $$f:X\to \omega$$, where $$\omega$$ is the set of natural numbers. Uncountable means not countable.

By a continuum I shall mean a connected compact Hausdorff space.

Theorem (ZF). Every non-degenerate continuum is uncountable.

Proof. Let $$X$$ be a non-degenerate continuum.

For a contradiction suppose $$X$$ is countable. Apparently $$X$$ must be infinite, and so we may enumerate $$X=\{x_i:i<\omega\}$$ where the $$x_i$$'s are distinct.

Let $$C_0=X$$.

Suppose $$n\geq 1$$ and non-degenerate continua $$C_0\supseteq C_1\supseteq ... \supseteq C_{n-1}$$ have been defined.

Let $$x^*$$ be the element of $$C_{n-1}$$ with least subscript.

Let $$x^{**}$$ be the element of $$C_{n-1}$$ with least subscript greater than $$x^*$$'s.

Let $$\mathcal U_n=\{U\subseteq X:U \text{ is open, }x^*\in U\text{, and }x^{**}\notin \overline U\}.$$ Since $$X$$ is Hausdorff, $$\mathcal U_n\neq\varnothing$$. Let $$\mathcal C_n=\{C\subseteq C_{n-1}:x^{**}\in C,\;C\text{ is connected, and }(\exists U\in \mathcal U_n)(C\cap U=\varnothing)\}.$$Let $$C_n=\overline{\bigcup \mathcal C_n}$$. Then $$C_n$$ is a continuum, and is non-degenerate because some elements of $$\mathcal C_n$$ are non-degenerate. This is true because compactness and normality of $$X$$ implies the quasi-component of $$x^{**}$$ in $$C_{n-1}\setminus U$$, $$U\in \mathcal U_n$$, is connected, and this quasi-component must meet $$\partial U$$ in order for $$X$$ to be connected. (Choice is not needed to prove normality, nor is it needed to prove these quasi-components are connected.)

Continuing in this manner, we construct a nested sequence $$(C_n)$$ of non-empty compact sets. Their intersection must be non-empty. But on the other hand we ensured each point of $$X$$ is eventually not in $$C_n$$. Contradiction. $$\blacksquare$$

Related: In 2013 Horst Herrlich and Kyriakos Keremedi proved that "Connected separable metric spaces need not have continuum size in ZF".

My follow-up question: Is every separable metric continuum equinumerable with the reals, in ZF?

• Dec 19 '18 at 2:47
• If you found a mistake in the answer, that information goes on top, not on the bottom. It is not a footnote announcement. Dec 19 '18 at 4:55
• By the way, going over the chat linked here (skimming is more appropriate), let me point out that the equivalence of "infinite" with "has a countably infinite subset" is in fact weaker than countable choice, which is weaker than dependent choice. But it still requires some choice nonetheless. Dec 19 '18 at 5:14
• Not sure. It is possible that a countable set has a Dedekind-finite collection of subsets, which makes this construction a bit iffy, since it's about subsets and not about points. It requires further thought, that's for sure. Dec 19 '18 at 5:20
• @dfeuer: No choice is needed. If $A$ is a compact subset and $x\notin A$, then look at the collection of all open sets which separate a point in $A$ from $x$. It is an open cover of $A$, it has a finite subcover, which means that $x$ is not a limit point. Dec 19 '18 at 5:32