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Let's say I have an object which can be viewed as family of sets $\mathfrak{S} \subseteq 2^S$, and I want to prove that its intersection is non-empty. What is known already:

  1. $S$ is set of measurable functions $f : \Omega \rightarrow X$ on fixed probability space $\langle \Omega, \mathfrak{I}, \mathbb{P} \rangle$ with same finite codomain;
  2. Every finite sub-family of $\mathfrak{S}$ has non-empty intersection;
  3. Under metric $d(f', f'') = \mathbb{P}(f' \neq f'')$ every set in family is complete, but, unfortunately, not totally bounded.

Obviously, this is not enough to deduct non-emptiness of family intersection, but may be you can suggest some strategy? What additional facts can I try to prove here to reach my goal?

UPDATE: I have managed to prove fact in question. In my case every $S$ happened to be derivable through compactness-preserving means from another set of functions, which compactness can be proven in Tychonoff topology.

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One approach is to choose a weaker topology, where you might get compactness.

For instance, let's identify your finite set $X$ with $\{1,2,\dots,N\} \subset \mathbb{R}$. Then we can view your functions $f$ as elements of $L^2(\Omega, \mathbb{P})$ and equip this space with the weak topology. Since all your functions $f$ are bounded by $N$ and therefore have $\|f\|_{L^2} \le N$, by Alaoglu's theorem they form a precompact set in the weak topology. So if you could show that your sets $S_i \in \mathfrak{S}$ are weakly closed in $L^2$, they will be weakly compact and you will have your desired result.

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  • $\begingroup$ OK, looks like I need some time to decipher this - I'm specializing in game theory, so stuff like this is usually a bit outside of my scope. $\endgroup$
    – Doktor Diagoras
    Commented Apr 5, 2019 at 17:15
  • $\begingroup$ I have been digging through Alaoglu's theorem and noticed a thing that can probably save me from going full $L^2$. It happened that all my $S_i \in \mathfrak{S}$ have nice property - if $f' \stackrel{\mathbb{P}}{=} f''$ then $f' \in S_i \Leftrightarrow f'' \in S_i$, or, speaking simply, if function belongs to some set, then such set contains all functions almost equal to it too. In such case, is it possible to prove its compactness through Tychonoff's theorem, or maybe I'm missing something? $\endgroup$
    – Doktor Diagoras
    Commented Apr 5, 2019 at 19:56
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    $\begingroup$ @DoktorDiagoras: Well, it's good that your sets are closed under a.e. equality, otherwise it wouldn't be well defined to treat your functions as elements of $L^2$. I don't see that it particularly helps you avoid $L^2$. Alaoglu is really not much more than Tychonoff with window dressing. The real issue is coming up with a topology for which Tychonoff would be useful, i.e. which embeds into some product topology. The advantage of the weak topology is that this has already been done for you. $\endgroup$
    – Nate Eldredge
    Commented Apr 5, 2019 at 20:25
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    $\begingroup$ "Closed under a.e. equality" does not imply that you have closed sets in the product topology. Consider for example $\Omega = \mathbb{N}$ with $\mathbb{P}(\{n\}) = 2^{-n}$. The only null set is empty so your "closed under a.e. equality" is trivial. This does not imply that every set is closed in the product topology. $\endgroup$
    – Nate Eldredge
    Commented Apr 5, 2019 at 20:57
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    $\begingroup$ If you're able to prove that your sets $S_i$ are closed in the product topology, great. My guess is that you will find this difficult or impossible; the product topology is just not well suited to such problems. For instance, it is possible for a set to consist only of measurable functions, yet its closure contains functions which are not measurable. So one of the immediate benefits of $L^2$ is that it only has measurable functions to begin with. $\endgroup$
    – Nate Eldredge
    Commented Apr 5, 2019 at 20:59

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