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.


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 $X$.

  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $\kappa$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.