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A parallel metrization theorem

  • Taras BanakhEmail author
  • Olena Hryniv
Open Access
Research Article
  • 42 Downloads

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 Open image in new window 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 Open image in new window is open in X;

  • upper semicontinuous if for any closed set \(F\subset X\) its \({\mathscr {C}}\)-star Open image in new window 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 Open image in new window . To see that Open image in new window is open, take any point Open image in new window 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 Open image in new window . 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 Open image in new window .

To see that \({\mathscr {C}}\) is upper semicontinuous, fix any closed set \(F\subset X\) and consider its \({\mathscr {C}}\)-star Open image in new window . To see that Open image in new window is closed, take any point Open image in new window and find a set \(C\in {\mathscr {C}}\) such that \(s\in C\). It follows from Open image in new window that \(C\cap F=\varnothing \) and hence Open image in new window by the compactness of C. We claim that Open image in new window . Assuming the opposite, we can find a point Open image in new window 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 Open image in new window of the locally finite cover \({\mathscr {V}}\). Since the cover \({\mathscr {C}}\) is upper semicontinuous, the set Open image in new window 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 Open image in new window is open and hence Open image in new window is an open neighborhood of C.

Put Open image in new window 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 \(F_C\). HenceNext, 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 \(x,y\in X\) letAdjust the function \(\delta \) to a pseudometric d lettingwhere 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 thatwhich is a desired contradiction. By analogy we can prove that the case \(d(A,B)<d(A,b)\) leads to a contradiction. \(\square \)

Notes

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)zbMATHGoogle Scholar

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© The Author(s) 2019

OpenAccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  1. 1.Institute of MathematicsJan Kochanowski University in KielceKielcePoland
  2. 2.Faculty of Mechanics and MathematicsIvan Franko National University of LvivL’vivUkraine

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