A parallel metrization theorem

Abstract

Two non-empty sets A, B of a metric space (X, d) are called parallel if $d(a,B)=d(A,B)=d(A,b)$ for any points $a\in A$ and $b\in B$. Answering a question posed on mathoverflow.net, we prove that for a cover ${\mathscr {C}}$ of a metrizable space X by compact subsets, the following conditions are equivalent: (i) the topology of X is generated by a metric d such that any two sets $A,B\in {\mathscr {C}}$ are parallel; (ii) the cover ${\mathscr {C}}$ is disjoint, lower semicontinuous and upper semicontinuous.

In this paper we shall prove a “parallel” metrization theorem answering a question [1] of the Mathoverflow user116515. The question concerns parallel sets in metric spaces.

Two non-empty sets A, B in a metric space (X, d) are called parallel if

$$\begin{aligned} d(a,B)=d(A,B)=d(A,b)\quad \hbox {for any } a\in A \hbox { and } b\in B. \end{aligned}$$

Here $d(A,B)=\inf \{ d(a,b):a\in A,\, b\in B\}$ and for $x\in X$. Observe that two closed parallel sets A, B in a metric space are either disjoint or coincide.

Let ${\mathscr {C}}$ be a family of non-empty closed subsets of a topological space X. A metric d on X is defined to be ${\mathscr {C}}$-parallel if any two sets $A,B\in {\mathscr {C}}$ are parallel with respect to the metric d.

A family ${\mathscr {C}}$ of subsets of X is called a compact cover of X if $X=\bigcup {\mathscr {C}}$ and each set $C\in {\mathscr {C}}$ is compact.

FormalPara Problem

([1]) For which compact covers ${\mathscr {C}}$ of a topological space X the topology of X is generated by a ${\mathscr {C}}$-parallel metric?

A metric generating the topology of a given topological space will be called admissible. A necessary condition for the existence of an admissible ${\mathscr {C}}$-parallel metric is the upper and lower semicontinuity of the cover ${\mathscr {C}}$.

A family ${\mathscr {C}}$ of subsets of a topological space X is called

• lower semicontinuous if for any open set $U\subset X$ its ${\mathscr {C}}$-star is open in X;

• upper semicontinuous if for any closed set $F\subset X$ its ${\mathscr {C}}$-star is closed in X;

• continuous if ${\mathscr {C}}$ is both lower and upper semicontinuous;

• disjoint if any distinct sets $A,B\in {\mathscr {C}}$ are disjoint.

The following theorem is the main result of the paper, answering Problem.

FormalPara Theorem

For a compact cover ${\mathscr {C}}$ of a metrizable topological space X the following conditions are equivalent:

• (i) the topology of X is generated by a ${\mathscr {C}}$-parallel metric;

• (ii) the family ${\mathscr {C}}$ is disjoint and continuous.

FormalPara Proof

(i) $\Rightarrow$ (ii). Assume that d is an admissible ${\mathscr {C}}$-parallel metric on X. The disjointness of the cover ${\mathscr {C}}$ follows from the obvious observation that two closed parallel sets in a metric space are either disjoint or coincide.

To see that ${\mathscr {C}}$ is lower semicontinuous, fix any open set $U\subset X$ and consider its ${\mathscr {C}}$-star . To see that is open, take any point and find a set $C\in {\mathscr {C}}$ such that $s\in C$ and $C\cap U\ne \varnothing$. Fix a point $u\in U\cap C$ and find $\varepsilon >0$ such that the $\varepsilon$-ball $B(u;\varepsilon )=\{ x\in X:d(x,u)<\varepsilon \}$ is contained in U. We claim that . Indeed, for any $x\in B(s;\varepsilon )$ we can find a set $C_x\in {\mathscr {C}}$ containing x and conclude that $d(C_x,u)=d(C_x,C)\leqslant d(x,s)<\varepsilon$ and hence $C_x\cap U\ne \varnothing$ and .

To see that ${\mathscr {C}}$ is upper semicontinuous, fix any closed set $F\subset X$ and consider its ${\mathscr {C}}$-star . To see that is closed, take any point and find a set $C\in {\mathscr {C}}$ such that $s\in C$. It follows from that $C\cap F=\varnothing$ and hence by the compactness of C. We claim that . Assuming the opposite, we can find a point and a set $C_x\in {\mathscr {C}}$ such that $x\in C_x$ and $C_x\cap F\ne \varnothing$. Fix a point $z\in C_x\cap F$ and observe that $d(C,F)\leqslant d(C,z)=d(C,C_x)\leqslant d(s,x)<\varepsilon =d(C,F)$, which is a desired contradiction.

(ii) $\Rightarrow$ (i). The proof of this implication is more difficult. Assume that ${\mathscr {C}}$ is disjoint and continuous. Fix any admissible metric $\rho \leqslant 1$ on X.

Let ${\mathscr {U}}_0(C)=\{X\}$ for every $C\in {\mathscr {C}}$.

FormalPara Claim

For every $n\in {\mathbb {N}}$ and every $C\in {\mathscr {C}}$ there exists a finite cover ${\mathscr {U}}_n(C)$ of C by open subsets of X such that

• (a) each set $U\in {\mathscr {U}}_n(C)$ has $\rho$-diameter $\leqslant 1/{2^n}$;

• (b) if a set $A\in {\mathscr {C}}$ meets some set $U\in {\mathscr {U}}_n(C)$, then $A\subset \bigcup {\mathscr {U}}_n(C)$ and A meets each set $U'\in {\mathscr {U}}_n(C)$.

FormalPara Proof

Using the paracompactness [2, 5.1.3] of the metrizable space X, choose a locally finite open cover ${\mathscr {V}}$ of X consisting of sets of $\rho$-diameter $<1/{2^n}$.

For every compact set $C\in {\mathscr {C}}$ consider the finite subfamily of the locally finite cover ${\mathscr {V}}$. Since the cover ${\mathscr {C}}$ is upper semicontinuous, the set is closed and disjoint with the set C. Since ${\mathscr {C}}$ is lower semicontinuous, for any open set $V\in {\mathscr {V}}(C)$ the set is open and hence is an open neighborhood of C.

Put and observe that ${\mathscr {U}}_n$ satisfies condition (a).

Let us show that the cover ${\mathscr {U}}_n(C)$ satisfies condition (b). Assume that a set $A\in {\mathscr {C}}$ meets some set $U\in {\mathscr {U}}_n(C)$. First we show that $A\subset \bigcup {\mathscr {U}}_n(C)$. Find a set $V\in {\mathscr {V}}(C)$ such that $U=W(C)\cap V$. It follows from $\varnothing \ne A\cap U\subset A\cap W(C)$ that the set A meets W(C) and hence is contained in W(C) and is disjoint with $F_C$. Hence

Next, take any set $U'\in {\mathscr {U}}_n(C)$ and find a set $V'\in {\mathscr {V}}(C)$ with $U'=W(C)\cap V'$. The relation $A\cap W(C)\cap V=A\cap U\ne \varnothing$ and the definition of the set $W(C)\supset A$ imply that A intersects the set $V'\in {\mathscr {V}}(C)$ and hence intersects the set $U'=W(C)\cap V'$. $\blacksquare$

Given two points $x,y\in X$ let

Adjust the function $\delta$ to a pseudometric d letting

where the infimum is taken over all sequences $x=x_0$, $\dots$, $x_m=y$. Condition (a) of Claim implies that $\rho (x,y)\leqslant \delta (x,y)$ and hence $\rho (x,y)\leqslant d(x,y)$ for any $x,y\in X$. So, the pseudometric d is a metric on X such that the identity map $(X,d)\rightarrow (X,\rho )$ is continuous. To see that this map is a homeomorphism, take any point $x\in X$ and $\varepsilon >0$. Find $n\in {\mathbb {N}}$ such that $1/{2^n}<\varepsilon$ and choose a set $C\in {\mathscr {C}}$ with $x\in C$ and a set $U\in {\mathscr {U}}_n(C)$ with $x\in U$. Then for any $y\in U$ we get $d(y,x)\leqslant \delta (x,y)\leqslant 1/{2^n}<\varepsilon$, which means that the map $X\rightarrow (X,d)$ is continuous.

Finally, let us prove that the metric d is ${\mathscr {C}}$-parallel. Pick any two distinct compact sets $A,B\in {\mathscr {C}}$. We need to show that $d(a,B)=d(A,B)=d(A,b)$ for any $a\in A$, $b\in B$. Assuming that this inequality is not true, we conclude that either $d(a,B)>d(A,B)>0$ or $d(A,b)>d(A,B)>0$ for some $a\in A$ and $b\in B$.

First assume that $d(a,B)>d(A,B)$ for some $a\in A$. Choose points $a'\in A$, $b'\in B$ such that $d(a',b')=d(A,B)<d(a,B)$. By the definition of the distance $d(a',b')<d(a,B)$, there exists a chain $a'=x'_0, x'_1,\dots ,x'_m=b'$ such that $\sum _{ i=1}^{ m}\delta (x'_{i-1},x'_i)<d(a,B)$. We can assume that the points $x'_0,\dots ,x'_m$ are pairwise distinct, so for every $i\leqslant m$ there exist $n_i\geqslant 0$ such that $\delta (x'_{i-1},x'_i)= 1/{2^{n_i}}$ and hence $x'_{i-1},x'_i\in U_i'$ for some $C_i\in {\mathscr {C}}$ and $U_i'\in {\mathscr {U}}_{n_i}(C_i)$. For every $i\leqslant m$ let $A_i\in {\mathscr {C}}$ be the unique set with $x_i'\in A_i$. Then $A_0=A$ and $A_m=B$.

Using condition (b), we can inductively construct a sequence of points $a=x_0,x_1,\dots ,x_m\in B$ such that for every positive $i\leqslant m$ the point $x_i$ belongs to $A_i$ and the points $x_{i-1},x_i$ belong to some set $U_i\in {\mathscr {U}}_{n_i}(C_i)$. The chain $a=x_0,x_1,\dots ,x_m\in A_m=B$ witnesses that

which is a desired contradiction. By analogy we can prove that the case $d(A,B)<d(A,b)$ leads to a contradiction. $\square$