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?

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – dfeuer
    Dec 19 '18 at 2:47
  • 1
    $\begingroup$ If you found a mistake in the answer, that information goes on top, not on the bottom. It is not a footnote announcement. $\endgroup$
    – Asaf Karagila
    Dec 19 '18 at 4:55
  • 1
    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Dec 19 '18 at 5:14
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Dec 19 '18 at 5:20
  • 2
    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Dec 19 '18 at 5:32

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