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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?

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.

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 has non-empty intersection;
3. Under metric $d(f', f'') = \mathbb{P}(\{\omega \in \Omega \mid f'(\omega) \neq f''(\omega)\})$ 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?

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?