Lately, with Peter Scholze’s MathOverflow post about Grothendieck universes and the Isabelle/HOL implementation of schemes, it seems that in the sphere of online math there has been a somewhat renewed interest in when large cardinals make proving theorems easier. (Specifically, it is not necessary that one actually needs the large cardinals to prove the theorem — only that it makes the proof easier!) So I thought it would be fun to look through some old homework of mine and see if I could find an example where if I had allowed myself the use of a large cardinal, my life would have been easier. I found an example from when I took a course in C*-algebras a few years ago.

Let X be a locally compact Hausdorff space. By a compactification of X we mean an open dense embedding $X \to Y$ where Y is a compact Hausdorff space. By Alexandroff’s theorem, X always has a compactification, but in general if X is not compact then X may have multiple compactifications. We consider the category Comp X of compactifications of X equipped with continuous surjections which preserve X; the Alexandroff compactification is the final object of Comp X.

The Stone–Čech theorem. The category Comp X has an initial object.

One may show that the initial object of Comp X is $\text{Spec } C_b(X)$ where $C_b(X)$ is the Banach space of bounded continuous functions on X with its supremum norm, and the functor Spec is taken in the sense of C*-algebras (thus Spec A consists of maximal closed ideals equipped with the Zariski-Jacobson topology). This proof is presumably inoffensive to anyone who accepts ZFC (and offensive to anyone who does not, since one needs Zorn’s lemma to show that $C_b(X)$ has a maximal ideal in general — and ZF alone cannot prove that Comp X has an initial object).

However, for the purposes of the result I was trying to prove, I needed a proof of the Stone–Čech theorem that did not rely on the existence of $\text{Spec } C_b(X)$, or else my argument would have been circular. To do this, one proceeds as follows. If $Z \to Y$ is a morphism in Comp X, then since X is dense in Z, the underlying continuous surjection $Z \to Y$ is completely determined by its behavior on X, but it is also the identity on X. Therefore Comp X is a poset category. Let $\mathcal C$ be a chain in Comp X; then $\mathcal C$ is an inverse system of topological spaces, and if C is the inverse limit of $\mathcal C$, then one can show that there is a closed embedding $C \to \prod \mathcal C$. Since $\prod \mathcal C$ is a compact Hausdorff space by Tychonoff’s theorem, so is C. Taking the inverse limits of the open dense embeddings $X \to Y$, where $Y \in \mathcal C$, we obtain an open dense embedding $X \to C$, so C is an upper bound of $\mathcal C$ in Comp X.

At this point, one may proceed in two ways. Working in ZFC, it is only valid to apply Zorn’s lemma if Comp X is equivalent to a small category, but $\text{Comp } \mathbb N$ is a large category. To see that Comp X is equivalent to a small category, it suffices to show that there is a cardinal $\kappa$ such that every compactification of Comp X has at most $\kappa$ points; then for every compactification Y of X, one can find a compactification Z of X such that $Y \cong Z$ in Comp X, and the set-theoretic rank of Z is at most $\kappa$, and so Comp X is a subset of the set $V_\kappa$. Furthermore, if Y is a compactification of X and $y \in Y$, then, since X is dense in Y, by the boolean prime ideal theorem there is an ultrafilter U on the set Open X of open subsets of X such that $\lim U = y$. Since Y is Hausdorff, it follows that y is the UNIQUE limit of U, but some cardinal arithmetic can be used to show that if $\lambda$ is the cardinality of X, then there are only $2^{2^\lambda}$ ultrafilters on Open X (since elements of an ultrafilter on Open X are open subsets of X), so the cardinality of Y is at most $2^{2^\lambda}$. Therefore we may let $\kappa = 2^{2^{\lambda}}$.

Okay, that was stupid. We can also proceed by large cardinals. The following argument feels much more conceptual to me:

Definitions. Let $\delta > \aleph_0$ be a regular cardinal. We say that $\delta$ is an inaccessible cardinal if for every cardinal $\lambda < \delta$, $2^\lambda < \delta$. We say that $\delta$ is a hyperinaccessible cardinal if $\delta$ is an inaccessible cardinal and there is an increasing chain of inaccessible cardinals $\delta_\alpha$ such that $\lim_\alpha \delta_\alpha = \delta$.

Let $\delta$ be a hyperinaccessible cardinal and suppose that $\text{card }X < \delta$. Then there are inaccessible cardinals $\text{card }X < \kappa < \kappa' < \delta$. If $X \in V_\kappa$ and Y is a compactification of X, then Y can be obtained as an extension of the Alexandroff compactification by splitting nets, but $V_\kappa$ is a Grothendieck universe and so the topology of X can be already probed by nets in $V_\kappa$; therefore $Y \in V_\kappa$. Therefore $\text{Comp } X \subseteq V_\kappa$ is a small category in $V_{\kappa'}$, so X has a Stone–Čech compactification $\beta X$ with $\text{card } \beta X < \kappa' < \delta$.

This argument looks verbose, but only because I have written out the details; I think in practice I would just say that if X lies underneath an inaccessible cardinal $\kappa$, then enough nets to probe the topology of X are also under $\kappa$, so every compactification is as well.

1. James Leng says:

So stone cech compactification is not unique and depends on a measure given on X?

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1. Aidan Backus says:

No, I was being sloppy and wrote L^\infty norm to mean sup norm. I’ll correct that now…

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2. James Leng says:

Okay thanks. I thought I had read something incorrect about Stone-Cech compactifications in the past. This clarifies things.

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