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Say that a property $\mathscr{P}$ of topological spaces is incompressible iff there is some cardinal $\kappa$ such that, for all cardinals $\lambda$, there is a topological space $\mathcal{X}$ of size $>\lambda$ which has property $\mathscr{P}$ but has no subspace $\mathcal{Y}$ with property $\mathscr{P}$ and $\kappa\le \vert \mathcal{Y}\vert<\vert\mathcal{X}\vert$.

For example, connectedness is incompressible via $\kappa=2$: consider the Dedekind completion of a "large" dense linear order all of whose nondegenerate intervals have the same cardinality as the whole. On the other hand, hereditary properties like Hausdorffness are never incompressible since we can just look at any "moderately sized" subspace of a "large" appropriate space. Marginally less trivially, path connectedness is not incompressible since, unlike general connectedness, it is witnessed by "small configurations" (= paths).

My question is:

Is "compact + Hausdorff" incompressible?

As an aside, note that $\mathsf{ZFC}$ proves that $[0,1]$ has no uncountable compact subspaces of size $<2^{\aleph_0}$; consequently, if for example $2^{\aleph_0}=\aleph_2$ then $[0,1]$ is a space of size $\aleph_2$ with no compact subspace of size $\aleph_1$. So the set of sizes of compact spaces of a given space can be complicated.

Noah Schweber
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  • as a side note, if you're willing to work with theories rather than just sentences, it is not too hard to come up with a $C$-satisfiable theory $T$ none of whose $\models_C$-models are compact; see my comment here. (however, my example can only do this by adding a new constant symbol to the language; I wonder if it's possible in the pure language of rings.) – Atticus Stonestrom Aug 04 '21 at 06:48

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"Compact Hausdorff" is compressible. This follows from the fact that if $X$ is a Hausdorff space and $A\subseteq X$, then the closure of $A$ has cardinality at most $2^{2^{|A|}}$ (since every point in the closure is the limit of some ultrafilter on $A$). So, fixing $\kappa$, let $\lambda=2^{2^{\kappa}}$. If $X$ is any compact Hausdorff space of cardinality greater than $\lambda$, take a subset $A\subset X$ of size $\kappa$ and let $Y$ be the closure of $A$ in $X$. Then $Y$ is also compact Hausdorff and $\kappa\leq |Y|\leq\lambda<|X|$.

Eric Wofsey
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