# A parallel metrization theorem

• Taras Banakh
• Olena Hryniv
Open Access
Research Article

## Abstract

Two non-empty sets AB of a metric space (Xd) 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.

## Keywords

Metrization Parallel sets Metric space

## Mathematics Subject Classification

54E35

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 AB in a metric space (Xd) 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 AB 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.

### 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.

### Theorem

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

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

2. (ii)

the family $${\mathscr {C}}$$ is disjoint and continuous.

### 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 $$C\in {\mathscr {C}}$$.

### 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
1. (a)

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

2. (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 $$U'\in {\mathscr {U}}_n(C)$$.

### 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. $$\square$$

## References

1. 1.
user116515: Making compact subsets “parallel”. https://mathoverflow.net/questions/284544/making-compact-subsets-parallel. Accessed 27 Oct 2017
2. 2.
Engelking, R.: General Topology. Sigma Series in Pure Mathematics, vol. 6, 2nd edn. Heldermann, Berlin (1989)