# A Uniform Metric Selection Theorem

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.

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 $$F_y\subset f(q)\subset X\setminus G_y$$.