# “Indexed” version of compactness and Axiom of Choice

When thinking about some problems related to compactness, I thought of this notion which, at the first glance, seems similar to the usual definition of compactness.

A topological space is compact iff every open cover has a finite subcover. In the other words, if $$\{U_i; i\in I\}$$ is an open cover of $$X$$ then we have $$U_{i_1},\dots, U_{i_n}$$ such that $$U_{i_1}\cup\dots\cup U_{i_n}=X$$.

However, one should be a bit careful with the notation like this. An important thing to notice is that if the open cover is indexed by a set $$I$$, the index $$i\in I$$ needn't be uniquely determined by $$U$$. I.e., it is possible that $$U_i=U_j$$ for some $$i\ne j$$.

Indexed compactness. Let $$(X,\mathcal T)$$ be a topological space and $$f\colon I\to\mathcal T$$ be a function such that $$f[I]=\{f(i);i\in I\}$$ is an open cover of $$X$$. Then there is a finite set $$F\subseteq I$$ such that $$f[F]$$ is a cover of $$X$$.

The difference from the usual definition of compactness is that here when working with the finite subcover $$\{U_{i_1},\dots,U_{i_n}\}$$ we also have the indexes $$i_1,\dots,i_n$$. In the other words, if we have an open cover $$\{U_i; i\in I\}$$ (or a subcover $$\{U_i; i\in F\}$$, for an open set $$U$$ there might be several $$i$$'s such that $$U_i=U$$. This version of compactness selects one of them.

Of course, if we are working in ZFC, then using Axiom of Choice we can choose $$i\in I$$ for each open set in the cover (subcover). So in ZFC this is equivalent to the usual notion of compactness.

However, when we are are not allowed to use AC, my guess would be that this is no longer equivalent to the usual notion of compactness. (Or at least I do not see a straightforward way to prove the equivalence in ZF.)

Question. Is the "indexed-compactness" defined above equivalent to the usual definition of compactness (in ZF)? Was this version of compactness studied somewhere? Does equivalence between indexed-compactness and compactness imply AC?

I am aware that various versions of compactness are studied under ZF. Herrlich's book Axiom of Choice (Lecture Notes in Mathematics 1876) mentions filter-compact, ultrafilter-compact and Alexandroff-Urysohn-compact spaces. I haven't seen there something which would be equivalent to the above. (At least for none of the types of compactness defined there the relationship to "indexed-compactness" is immediately clear.)

• The indexing does not matter in the definition of compactness, any cover can be seen (choice-less) as an "indexed cover", etc. Engelking prefers to work with such indexed covers: it's a matter of style/presentation, IMHO. – Henno Brandsma Nov 18 at 16:27

It's fairly clear that 'indexed compactness' implies compactness. Conversely, we can turn any subset $$\mathcal{U} \subseteq \mathcal{T}$$ into an function $$f : I \to \mathcal{T}$$ by taking $$I = \mathcal{U}$$ and taking $$f$$ to be the inclusion map $$\mathcal{U} \hookrightarrow \mathcal{T}$$. For this choice of $$I$$ and $$f$$, the statement of 'indexed compactness' reduces to the usual statement of compactness.
• Your argument that any subset $\mathcal{U}\subseteq \mathcal{T}$ can be turned into a function $f\colon I\to \mathcal{T}$ is actually the argument that indexed compactness implies ordinary compactness. That is, in ordinary compactness we're given an open cover $\mathcal{U}\subseteq \mathcal{T}$, then we turn it into an indexed family (indexed by $\mathcal{U}$), apply indexed compactness, and obtain a finite subcover. The converse, that compactness implies indexed compactness, is what Martin is concerned about, and your answer doesn't address this. – Alex Kruckman Nov 18 at 16:56
• That is, the question is: Given an indexed cover $f\colon I\to \mathcal{T}$ (equivalently $(U_i)_{i\in I}$), we can apply compactness to obtain a finite subcover, but how can we extract the indices of this finite subcover (the $i_1,\dots,i_n$ in $I$ such that $\{U_{i_1},\dots,U_{i_n}\}$ is a cover) without choice. Of course, this is also easy to do, as explained in Andreas Blass's answer. – Alex Kruckman Nov 18 at 16:58