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Let $X$ and $Y$ be bounded complete separable metric spaces. Let $C = 2^\omega$ be Cantor space with its standard metric. All product spaces are taken to have the max metric.

Let $F, G \subseteq X\times Y$ be closed sets such that $\inf \{d(a,b) : a \in F, b \in G \} > 0$. For each $y\in Y$, let $F_y = \{ x\in X:(x,y) \in F \}$ and $G_y = \{ x \in X: (x,y) \in G\}$.

Does there always exist a set $Q \subseteq Y\times C$ and a function $f:Q \rightarrow 2^X$ such that

  • for every $y\in Y$ there exists a $c\in C$ such that $(y,c) \in Q$,
  • for every $q\in Q$, $f(q)$ is a closed subset of $X$,
  • $f$ is uniformly continuous in the Hausdorff metric, and
  • for every $(y,c) = q \in Q$, $F_y \subseteq f(q)$ and $f(q) \cap G_y = \varnothing$?

What if we let $C$ be Baire space, $\omega^\omega$, with its standard metric?

This is trivial when $Y$ is finite. It also always works when $X$ is compact. An extremely crass way to prove this is to let $g: C \rightarrow H(X)$ be a continuous surjection onto the hyperspace of $X$. Since this function is continuous, and therefore uniformly continuous, with regards to the Hausdorff metric on $H(X)$, the function $f(x,y,c)= g(c)$ is uniformly continuous on $X\times Y \times C$. For any given $y$ we can just choose a $c$ such that $g(c) = F_y$.

It's fairly easy to find examples--even when $X$ and $Y$ are both compact--that show that something like $C$ is necessary, i.e. if we take $C$ to be a one point space then there does not always exist such an $f$. For example if we take $X=Y=[0,1]$ and let $F = \{ (0,0),(1,1) \}$ and $G = \{(x,1-x):x \in [0,1]\}$, then there can be no such $f$ by topological considerations.

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For $C=\omega^\omega$ the answer is affirmative:

Theorem 1. Let $X$ be a metric space, $Y$ be separable metric space, and $F,G\subset X\times Y$ be closed sets such that $\inf\{d(x,y):x\in F,\;y\in G\}>0$. Then there exists a subset $Q\subset Y\times \omega$ and a function $f:Q\to 2^X$ such that

$\bullet$ for every $y\in Y$ there exists a unique $n\in\omega$ such that $(y,n)\in Q$;

$\bullet$ for every $q\in Q$, $f(q)$ is a closed subset of $X$;

$\bullet$ $f$ is uniformly continuous in Hausdorff metric;

$\bullet$ for every $(y,n)\in Q$ we have $F_y\subset f(q)\subset X\setminus G_y$.

Proof. Let $d_X,d_Y$ be the metrics of the spaces $X,Y$, and $d$ be the metric on $X\times Y$ defined by $d((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}$.

By our assumption, the real number $\varepsilon:=\inf\{d(a,b):a\in F,\;g\in G\}$ is positive. Let $\{Y_n\}_{n\in\omega}$ be a countable partition of the separable metric space $Y$ into sets of diameter $<\varepsilon$.

For every $n\in\omega$ let $E_n$ be the closure of the set $\bigcup_{y\in Y_n}F_y$ in $X$. We claim that $E_n\cap G_y=\emptyset$ for any $y\in Y_n$. To derive a contradiction, assume that $E_n\cap G_y$ contains some point $x$. Then $(x,y)\in G$. Since $x\in E_n$ belongs to the closure of $\bigcup_{z\in Y_n}F_z$, there exist $z\in Y_n$ and $x'\in F_z$ such that $d(x,x')<\varepsilon$. Now we see that $(x,z)\in F$ and $(x,y)\in G$ and $$\varepsilon \le d((x,y),(x',z))=\max\{d_X(x,x'),d_Y(y,z)\}<\varepsilon,$$which is a desired contradiction showing that $E_n\cap G_y=\emptyset$ for any $y\in Y_n$.

Now put $Q=\{(y,n)\in Y\times\omega:y\in Y_n\}$ and consider the map $f:Q\to 2^X$, $(y,n)\mapsto E_n$. It is easy to see that $f$ has the required properties.$\quad \square$

By analogy one can prove

Theorem 2. Let $X$ be metric space, $Y$ be a totally bounded metric space and $F,G\subset X\times Y$ be closed sets such that $\inf\{d(x,y):x\in F,\;y\in G\}>0$. Then there exists a finite metric space $M$, a subset $Q\subset Y\times M$ and a function $f:Q\to 2^X$ such that

$\bullet$ for every $y\in Y$ there exists a unique $z\in M$ such that $(y,z)\in Q$;

$\bullet$ for every $q\in Q$, $f(q)$ is a closed subset of $X$;

$\bullet$ $f$ is uniformly continuous in Hausdorff metric;

$\bullet$ for every $(y,z)\in Q\subset Y\times M$ we have $F_y\subset f(q)\subset X\setminus G_y$.

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