A parallel metrization theorem
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Abstract
Two nonempty 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.
Keywords
Metrization Parallel sets Metric spaceMathematics Subject Classification
54E35In 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.
Let \({\mathscr {C}}\) be a family of nonempty 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 \({\mathscr {C}}\).

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.
Theorem
 (i)
the topology of X is generated by a \({\mathscr {C}}\)parallel metric;
 (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
 (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 \(U'\in {\mathscr {U}}_n(C)\).
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).
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 \(\sum _{ i=1}^{ m}\delta (x'_{i1},x'_i)<d(a,B)\). 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 \(\delta (x'_{i1},x'_i)= 1/{2^{n_i}}\) and hence \(x'_{i1},x'_i\in U_i'\) 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 \(A_m=B\).
Notes
References
 1.user116515: Making compact subsets “parallel”. https://mathoverflow.net/questions/284544/makingcompactsubsetsparallel. Accessed 27 Oct 2017
 2.Engelking, R.: General Topology. Sigma Series in Pure Mathematics, vol. 6, 2nd edn. Heldermann, Berlin (1989)zbMATHGoogle Scholar
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