# How many compact Hausdorff spaces are there of a given cardinality?

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.

## 2 Answers

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 $X$.

• Nice. Alternatively, you can start by taking my functions $f_{xy}$ to get a continuous injection from $X$ to $[0,1]^\kappa$, which is then automatically an embedding by compactness. – Eric Wofsey Jul 30 '16 at 6:57

For any uncountable $\kappa$, there are exactly $2^\kappa$ compact Hausdorff spaces of cardinality $\kappa$, up to homeomorphism.

First, let's prove the upper bound (the following argument only requires $\kappa$ to be infinite). Suppose $X$ is a compact Hausdorff space of cardinality $\kappa$. For each pair of distinct points $x,y\in X$, choose a continuous function $f_{xy}:X\to\mathbb{C}$ such that $f_{xy}(x)\neq f_{xy}(y)$. Let $A\subseteq C(X)$ be the smalest $\mathbb{Q}[i]$-subalgebra containing the functions $f_{xy}$ and closed under taking complex conjugates and positive square roots (when they exist). Then $|A|=\kappa$. Note that the sup norm on $A$ can be defined using only complex conjugation and the $\mathbb{Q}[i]$-algebra structure: $\|f\|^2$ is the inf of all $r\in\mathbb{Q}$ such that $r-f\bar{f}$ has a square root which is its own conjugate.

By Stone-Weierstrass, $A$ is dense in $C(X)$, so we can recover $C(X)$ as the completion of $A$ with respect to the sup norm. The complex *-algebra structure of $C(X)$ can also be recovered from the $\mathbb{Q}[i]$-algebra structure on $A$ and the complex conjugation operation on $A$. By Gelfand duality, $X$ can be recovered up to homeomorphism from the complex *-algebra structure of $C(X)$.

Thus we can recover $X$ up to homeomorphism from the $\mathbb{Q}[i]$-algebra with conjugation $A$. But there are at most $2^\kappa$ structures of a $\mathbb{Q}[i]$-algebra with conjugation on a set of size $\kappa$. Thus there are at most $2^\kappa$ choices for $A$ up to isomorphism, so there are at most $2^\kappa$ such spaces $X$ up to homeomorphism.

Now we prove that there do exist $2^\kappa$ nonhomeomorphic compact Hausdorff spaces of cardinality $\kappa$ (as noted in the question, this statement is equivalent to CH if $\kappa=\aleph_0$, so we will need to use the uncountability of $\kappa$). The construction is very similar to my answer to this question asking how many metric spaces there are of a given cardinality, except that we can't use $\mathbb{Q}$ since we need the space to be compact. As a result, we need to find a different way to "distinguish" certain points of our space; we will do so using cofinality, which is where the requirement that $\kappa$ is uncountable comes in.

First let's make some definitions. If $X$ is a topological space, $x\in X$, and $\lambda$ is an infinite regular cardinal, let us say that $\lambda$ is a cofinality of $x$ if there exists a continuous injection $f:\lambda+1\to X$ such that $f(\lambda)=x$. Note, for instance, if $X$ is an ordinal and $x\in X$ is a limit ordinal, then $x$ has only one cofinality in this sense and it agrees with the usual notion of the cofinality of $x$. More generally, if $X$ is a Dedekind-complete totally ordered set with the order topology, then $\lambda$ is a cofinality of $x$ iff $\lambda$ is either the cofinality of $x$ from below or the cofinality of $x$ from above, in the order-theoretic sense.

Recall also the notion of Cantor-Bendixson rank. If $X$ is a topological space, write $D(X)$ for the set of non-isolated points of $X$ (the "Cantor-Bendixson derivative of $X$"). For each ordinal $\alpha$, define $D^\alpha(X)$ by induction:

• $D^0(X)=X$

• $D^{\alpha+1}(X)=D(D^\alpha(X))$

• For $\alpha$ limit, $D^\alpha(X)=\bigcap_{\beta<\alpha} D^\beta(X)$.

For $x\in X$, the least $\alpha$ such that $x\not\in D^{\alpha+1}(X)$ is called the Cantor-Bendixson rank of $x$ (if any such $\alpha$ exists). If $X$ is an ordinal and $x\in X$ is the ordinal $\omega^\alpha$ for some ordinal $\alpha$, then it is well-known that $x$ has Cantor-Bendixson rank $\alpha$.

We are now ready to construct our $2^\kappa$ different compact Hausdorff spaces. If $S$ is any set of ordinals, let $\alpha=\sup S +1$ and let $L_S$ be the totally ordered set obtained from $\alpha$ by inserting a copy of $\omega^*$ (the natural numbers with the reverse ordering) between $\beta$ and $\beta+1$ for each $\beta\in S$. It is easy to see that $L_S$ is always Dedekind-complete and bounded, and is thus compact in the order topology.

Now let $T$ be any set of ordinals with the following properties:

• $|T|=\kappa$, and every element of $T$ has cardinality $\kappa$.

• Every element of $T$ has uncountable cofinality.

It is easy to see that there are $2^\kappa$ such sets $T$ (here is where we use that $\kappa$ is uncountable). Given such a $T$, let $S=\{\omega^\alpha:\alpha\in T\}$ and consider the space $L_S$. Then $L_S$ is a compact Hausdorff space of cardinality $\kappa$; I claim we can recover the set $T$ from $L_S$. Each point $\omega^\alpha$ for $\alpha\in T$ has two different cofinalities in $L_S$: its usual cofinality as an ordinal, which is uncountable, and $\omega$, because of the copy of $\omega^*$ we inserted just above $\omega^\alpha$. Moreover, these are the only points of $L_S$ which have more than one cofinality. Thus we can say that $T$ is the set of all ordinals $\alpha$ such that there exists a point of $L_S$ of Cantor-Bendixson rank $\alpha$ which has more than one cofinality.

Thus the spaces $L_S$ are nonhomeomorphic for different sets $T$, giving $2^\kappa$ nonhomeomorphic compact Hausdorff spaces of cardinality $\kappa$.