[Math] How many compact Hausdorff spaces are there of a given cardinality

Question

This is a question I found myself wondering about recently. I eventually figured out the answer myself, but as this doesn't seem to be written down anywhere easy to find on the Internet I decided to share it here.

Let $\kappa$ be an uncountable cardinal. How many compact Hausdorff spaces of cardinality $\kappa$ are there, up to homeomorphism?

It is fairly well-known that if $\kappa$ is an infinite cardinal, then there are $2^{2^\kappa}$ different topological spaces of cardinality $\kappa$, up to homeomorphism (see What is the cardinality of the set of all topologies on $\mathbb{R}$?, for instance). On the other hand, it is also fairly well-known that there are only $\aleph_1$ countable compact Hausdorff spaces up to homeomorphism (since they are all homeomorphic to countable ordinals; see Countable compact spaces as ordinals, for instance). So it's not obvious what to expect the answer to the question above to be.

Answer 1

As a matter of taste I prefer an argument more elementary than Eric Wofsey’s to show that there are at most $2^\kappa$ compact Hausdorff spaces of cardinality $\kappa$.

Lemma. Let $X$ be a compact Hausdorff space of cardinality $\kappa$; then $\chi(X)\le\kappa$, i.e., each point has a local base of cardinality at most $\kappa$.

Proof. Fix $x\in X$. For each $y\in X\setminus\{x\}$ there are disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. Let $\mathscr{U}=\big\{U_y:y\in X\setminus\{x\}\big\}$, and let $$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{U}\text{ and }\mathscr{F}\text{ is finite}\right\}\;.$$ Let $U$ be an open nbhd of $x$; $\{V_y:y\in X\setminus U\}$ is an open cover of the compact set $X\setminus U$, so there is a finite $F\subseteq X\setminus U$ such that $\{V_y:y\in F\}$ covers $U$. It follows that $\bigcap_{y\in F}U_y\subseteq U$ and hence that $\mathscr{B}$ is a local base at $x$. $\dashv$

It’s an immediate corollary that $w(X)\le|X|\cdot\chi(X)\le\kappa$, i.e., that $X$ has a base of cardinality $\kappa$.

$X$ is Tikhonov of weight at most $\kappa$, so $X$ can be embedded in the cube $[0,1]^\kappa$ of weight $\kappa$. $[0,1]^\kappa$ has at most (in fact exactly) $2^\kappa$ open subsets, so it has at most $2^\kappa$ closed subsets. $X$ embeds as one of these subsets, so there are at most $2^\kappa$ possibilities for $X$.