# When is the generalized Cantor space $\kappa$-compact?

My M.Sc. student has the following question, that I assume has an answer in the literature, and we are looking for references.

The generalized Cantor space is the space $2^\kappa$, with basic open sets $$[\sigma] := \{f\in 2^\kappa : \sigma\subseteq f\},$$ for $\sigma\in 2^{<\kappa}$.

A space is $\kappa$-compact if every open cover has a subcover of cardinality (strictly) smaller than $\kappa$.

For which cardinals $\kappa$ is the generalized Cantor space $2^\kappa$ $\kappa$-compact? We are interested in strong limit cardinals $\kappa$, mainly strongly inaccessible $\kappa$.

We would especially appreciate references.

Update: A related question.

• If my memory serves me right this is strongly compact cardinals (I'd say weakly compact, but I am going to hedge my bets here, and claim my memory says in the weakly compact case it is only true for covers of cardinality $\kappa$). But I've got a mild case of the flu, so there's no reason to trust my memory. – Asaf Karagila Mar 7 '16 at 16:59
• @Asaf: what about this argument by contradiction; for each $\alpha<\kappa$ the basic opens $[\sigma]$ with $\sigma \in 2^{\alpha}$ is an open covering of cardinality less than $\kappa$, so for at least one $\sigma$ the open $[\sigma]$ is not contained in the original open covering. Then by weak compactness there is a cofinal branch made of these $\sigma$'s which cannot be covered by the original open cover. – godelian Mar 7 '16 at 17:14
• @AsafKaragila: Be healthy, we need you here. :) Note that every open cover of $2^\kappa$ is refined by one with basic open sets. We assume $2^{<\kappa}=\kappa$, so we may restrict attention to open covers of cardinality $\kappa$. – Boaz Tsaban Mar 7 '16 at 17:36
• @godelian: You mean you are applying the tree property to the tree induced by all these $\sigma$-s? How do you prove the branch is not covered? In any case, I think I see why the tree property suffices, but with a slightly more involved argument. And what about the converse implication, why is it necessary? – Boaz Tsaban Mar 7 '16 at 17:44
• Boaz: Looks like something I should have noticed. So I will unhedge my bet, and go with weakly compact. @godelian: I'm afraid I don't have the mental capacity right now to think about this sort of mathematics. – Asaf Karagila Mar 7 '16 at 17:45

A cardinal $\kappa$ is weakly compact if and only if $2^{\kappa}$ is $\kappa$-compact and $\kappa$ is strongly compact if and only if $2^{I}$ is $\kappa$-compact for all sets $I$ where $2^{I}$ is given the topology with basis of open sets of the form $[\sigma]$ where $\sigma:J\rightarrow 2$ and $|J|<\kappa$.
To prove the tree property from the Tychonoff theorem characterization of weak compactness, one uses the following argument. Suppose that $T$ is a $\kappa$-tree. Then for each $\alpha<\kappa$, let $T_{\alpha}$ be the $\alpha$-th level of this tree. Then by a compactness argument, one can show that the inverse limit $\varprojlim_{\alpha<\kappa}T_{\alpha}\subseteq\prod_{\alpha<\kappa}T_{\alpha}$ is a non-empty closed subset (the proof of this fact is the same in the ordinary compactness case). However, $\varprojlim_{\alpha<\kappa}T_{\alpha}$ is the set of all $\kappa$-branches of $T$. Therefore $\kappa$ satisfies the tree property.
To prove that every weakly compact cardinal satisfies the $\kappa$-Tychonoff theorem, one could use the tree property and inaccessibility to show that every $\kappa$-filter on a $\kappa$-algebra of sets can be extended to a $\kappa$-ultrafilter and from this fact one could show that weakly compact cardinals satisfy a $\kappa$-version of Alexander's subbase theorem. From Alexander's subbase theorem, one then establishes that weakly compact cardinals satisfy the $\kappa$-Tychonoff theorem.