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.



Proving that family of sets has non-empty intersection
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.


