Let $X$ be a compact metric space. Say that two compact subsets $E,F\subset X$ are *parallel* if
$$ dist(x,F) = dist(y,E)$$
for all $x\in E$ and $y\in F$. Here $ dist(y,E) = \inf\{d(y,z):z\in E\}.$


The overall question I would like to understand is the following:



> 
> Let $X$ be a compact (Hausdorff, second countable, hence metrizable) topological space with a collection of disjoint compact subsets $\{E\_t\}\_{t\in T}$. When is there a metric on $X$ that makes every pair of sets in $\{E\_t\}$ parallel?
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I generally think of $T$ as uncountable and $\{E\_t\}$ as a partition of $X$, but I don't see the need to assume this.


There is at least one clear necessary condition that $\{E\_t\}$ must satisfy: Namely, if $\{x\_n\in E\_{t\_n}\}$ is a sequence converging to $x\in E\_t$, then every element $x'\in E\_t$ is the limit of a sequence $\{x'\_n\in E\_{t\_n}\}$, **and** every sequence $\{x''\_n\in E\_{t\_n}\}$ accumulates only at points in $E\_t$.


Call this condition **(\*)**. (The part after the **and** in condition (\*) was added after MTyson's example.)


For example, if $X=[0,1]$ and $\{E\_t\}$ consists of $[0,1/2]$ as well as all the singletons $\{x\}$ for each $x\in (1/2,1]$, the collection fails condition (\*) and can't be made parallel.


So I refine the question:



> 
> Let $X$ be a compact (Hausdorff, second countable, hence metrizable) topological space with a collection of disjoint compact subsets $\{E\_t\}\_{t\in T}$, satisfying (\*). Is there a metric on $X$ that makes every pair of sets in $\{E\_t\}$ parallel?
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As I don't know if I believe this, I also ask:



> 
> Are there conditions, in addition to or instead of (\*), that ensure the sets $\{E\_t\}$ can be made parallel by the appropriate choice of a metric on $X$?
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I'd be interested in counterexamples, other possible obstructions, references to similar things in the literature, or general sufficient conditions (even if they assume more about $X$ or $\{E\_t\}$).


One could imagine cases in which $T$ is a topological space and $t\mapsto E\_t$ is continuous in some topology on the space of compacta of $X$ (e.g., that induced by the Hausdorff metric coming from some metric on $X$.) Maybe this is equivalent to my formulation. Regardless, if that makes the problem easier or more well-known then by all means assume it.



Making compact subsets "parallel"
The answer to this problem is affirmative (at least for covers).


**Definition.** A family $\mathcal C$ of subsets of a topological space $X$ is called


$\bullet$ *lower semicontinuous* if for any open set $U\subset X$ its $\mathcal C$-star $St(U;\mathcal C):=\bigcup\{C\in\mathcal C:C\cap U\ne\emptyset\}$ is open in $X$;


$\bullet$ *upper semicontinuous* if for closed set $F\subset X$ its $\mathcal C$-star $St(F;\mathcal C)$ is closed in $X$;


$\bullet$ *continuous* if $\mathcal C$ is both lower and upper semicontinuous.


**Theorem.** For a cover $\mathcal C$ of a metrizable space $X$ by compact subsets the following conditions are equivalent:


1. The topology of $X$ is generated by a metric $d$ making all sets in the cover $\mathcal C$ parallel.
2. The cover $\mathcal C$ is disjoint and continuous.
3. The cover $\mathcal C$ has property $(\*)$.


*Proof*. The implications $(1)\Rightarrow(2)\Leftrightarrow(3)$ are simple and are left to the reader. 


To prove that $(2)\Rightarrow(1)$, assume $\mathcal C$ is disjoint and continuous. Fix any admissible metric $\rho\le 1$ on $X$.


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


**Claim.** For every $n\in\mathbb N$ and every $C\in\mathcal C$ there exists a finite cover $\mathcal U\_n(C)$ of $C$ by open subsets of $X$ such that


(i) each set $U\in\mathcal U\_n(C)$ has $\rho$-diameter $\le\frac1{2^n}$;


(ii) if a set $A\in\mathcal C$ meets some set $U\in\mathcal U\_n(C)$, then $A\subset\bigcup\mathcal U\_n(C)$ and $A$ meets each set $U'\in\mathcal U\_n(C)$.


*Proof of Claim.* Using the paracompactness of the metrizable space $X$, choose an open locally finite cover $\mathcal V$ of $X$ consisting of sets of $\rho$-diameter $<\frac1{2^n}$.


For every compact set $C\in\mathcal C$ consider the finite subfamily $\mathcal V(C):=\{V\in\mathcal V:V\cap C\ne \emptyset\}$ of the locally finite cover $\mathcal V$. Since the cover $\mathcal C$ is upper semicontinuous, the set $F\_C=St(X\setminus \bigcup\mathcal V(C);\mathcal C)$ is closed and disjoint with the set $C$. Since $\mathcal C$ is lower semi-continuous, for any open set $V\in\mathcal V(C)$ the set $St(V;\mathcal C)$ is open and hence $W(C):=\bigcap\_{V\in\mathcal V(C)}St(V;\mathcal C)\setminus F\_C$ is an open neighborhood of $C$.


It can be shown that the family $\mathcal U\_n(C):=\{W(C)\cap V:V\in\mathcal V(C)\}$ satisfies the conditions (i), (ii).


*Continuation of the proof of Theorem.* 
Given two points $x,y\in X$ let $$\delta(x,y):=\inf\big\{\tfrac1{2^n}:\mbox{$\exists C\in\mathcal C$ and $U\in\mathcal U\_n(C)$ such that $x,y\in U$}\big\}.$$
Adjust the function $\delta$ to a pseudometric $d$ letting $$d(x,y)=\inf\sum\_{i=1}^m\delta(x\_{i-1},x\_i)$$where the infimum is taken over all sequences $x=x\_0,\dots,x\_m=y$.


It can be shown that the metric $d$ generates the topology of $X$ and $\rho\le d$.


Let us prove any sets $A,B\in\mathcal C$ are parallel in the metric $d$. 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 equality 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 $i\le m$ there exist $n\_i\ge 0$ such that $\delta(x'\_{i-1},x'\_i)=\frac1{2^{n\_i}}$ and hence $x'\_{i-1},x'\_i\in U\_i'$ for some $C\_i\in\mathcal C$ and $U\_i'\in\mathcal U\_{n\_i}(C\_i)$. For every $i\le m$ let $A\_i\in\mathcal C$ be the unique set with $x\_i'\in A\_i$. Then $A\_0=A$ and $A\_m=B$.


Using the condition (ii), we can inductively construct a sequence of points $a=x\_0,x\_1,\dots,x\_m\in B$ such that for every positive $i\le m$ the point $x\_i$ belongs to $A\_i$ and the points $x\_{i-1},x\_i$ belong to some set $U\_i\in\mathcal U\_{n\_i}(C\_i)$. The chain $a=x\_0,x\_1,\dots,x\_m\in A\_m=B$ witnesses that $$d(a,B)\le d(a,x\_m)\le\sum\_{i=1}^m\delta(x\_{i-1},x\_i)\le\sum\_{i=1}^m\tfrac1{2^{n\_i}}=
\sum\_{i=1}^m\delta(x'\_{i-1},x\_i')<d(a,B),$$
which is a desired contradiction.


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


