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?

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?

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 $X$?

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.

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    $\begingroup$ For example: if the $E_t$ are basic clopen subsets of the Cantor space $2^\omega$, then the standard metric dist(x,y) = $1/2^{\min\{n : x(n) \neq y(n)\}}$ makes them parallel. I suppose, a bit more generally, if $X$ is the Cantor space and each $E_t$ is either a basic clopen set or a singleton, then this metric makes them all parallel. $\endgroup$
    – Will Brian
    Oct 27, 2017 at 18:41
  • $\begingroup$ Yes, nice example. $\endgroup$
    – user116515
    Oct 27, 2017 at 18:47
  • $\begingroup$ Are you assuming everywhere that $X$ is metrizable? $\endgroup$
    – MTyson
    Oct 27, 2017 at 19:01
  • $\begingroup$ Yes, good question. I should have been more precise -- I do certainly want to assume $X$ is metrizable (Hausdorff and second countable). I don't have very exotic topologies in mind. I will edit the question. $\endgroup$
    – user116515
    Oct 27, 2017 at 19:09

3 Answers 3


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.


No, (the old version of) $(*)$ alone doesn't imply that there is a metric making the sets parallel.

Take $X=[-1,1]$ and let $\{E_t\}=\{-t,t\}_{0\le t<1}\cup\{-1\}\cup\{1\}$. It's easy to see that $(*)$ is satisfied here. Suppose there were a metric making $d$ making the sets parallel. Then $d(\frac{n-1}{n},1)=d(-\frac{n-1}{n},1)$ yet $\frac{n-1}{n}\to 1$ and $-\frac{n-1}{n}\to -1$, so $d(-1,1)=0$.

To get around this counterexample, you could try assuming the leaves are connected or that $T$ (with the quotient topology?) is Hausdorff.

  • $\begingroup$ You are absolutely right. This suggests to me a small modification to (*) that your example lacks. If it is not gauche to do so, I will edit the question. $\endgroup$
    – user116515
    Oct 27, 2017 at 23:32
  • $\begingroup$ Condition (*) has no been slightly modified in light of your example, making it more symmetric. In particular, your example doesn't satisfy the new condition (*). Thanks for pointing this out; this was already helpful to me. $\endgroup$
    – user116515
    Oct 27, 2017 at 23:38

If $(E_t)_{t\in T}$ is a partition of a metric space $(X,d)$ into parallel compact subsets, then the metric $d$ induces a quotient metric $\hat d$ on the index set $T$. This metric is defined by $\hat d(t,s)=\inf\{d(x,y):x\in E_t,\;y\in E_s\}$ for $t,s\in T$.

It is easy to show that the quotient map $q:X\to T$ (defined by $q^{-1}(t)=E_t$) is open and perfect. We recall that a map $f:X\to Y$ is perfect if it is continuous, closed and has compact preimages of points.

So, in case of partitions, the original problem reduces to the following one:

Problem. Let $f:X\to Y$ be an open perfect map between metrizable spaces. Is the topology of $X$ generated by a metric making the fibers of the map $f$ parallel?

For zero-dimensional space $X$ the affirmative answer to this problem is given by the following

Theorem 0. For any open perfect map $f:X\to Y$ between metrizable spaces the space $X$ admits an ultrametric $d$ that generates the topology of $X$ and makes the fibers $f^{-1}(y)$, $y\in Y$, of the map $f$ parallel.

Proof of Theorem 0. Fix any metric $\rho$ generating the topology of the space $X$. Let $\mathcal U_0=\{X\}$.

Claim. There exists a sequence $(\mathcal U_n)_{n=0}^\infty$ of disjoint open covers of $X$ such that $\mathcal U_0=\{X\}$ and for every $n\in\mathbb N$ the following conditions are satisfied:

  1. Each $U\in\mathcal U_n$ is contained in some $V\in\mathcal U_{n-1}$;

  2. $\max_{U\in\mathcal U_n}diam_\rho(U)<\frac1n$;

  3. For any points $y,z\in Y$ if $f^{-1}(z)$ intersects some set in the family $\mathcal U_n(y):=\{U\in\mathcal U_n:U\cap f^{-1}(y)\ne\emptyset\}$, then $f^{-1}(z)$ intesects each set in the family $\mathcal U_n(y)$.

Proof of Claim. The Claim is proved by induction. Assume that for some $n\in\mathbb N$ the disjoint cover $\mathcal U_{n-1}$ has been constructed. Since the space $X$ is zero-dimensional, there exists a disjoint open cover $\mathcal V$ of $X$ such that each set $V\in\mathcal V$ has $\rho$-diameter $<\frac1n$ and is contained in some set $U\in\mathcal U_{n-1}$. For every $y\in Y$ consider the family $\mathcal V_y=\{V\in\mathcal V:V\cap f^{-1}(y)\ne\emptyset\}$, which is finite by the compactness of the fiber $f^{-1}(y)$. Since the map $f$ is open and closed, there exists a clopen neighborhood $W_y\subset Y$ of $y$ such that for each $z\in W_y$ the set $f^{-1}(z)$ is contained in $\bigcup \mathcal V_y$ and intersects each set $V\in\mathcal V_y$. Such neighborhood $W_y$ can be defined as $W_y:= \bigcap_{V\in\mathcal V_y}f(V)\setminus f(X\setminus\bigcup\mathcal V_y)$. Since the space $Y$ is zero-dimensional, there exists a disjoint open cover $\mathcal W$ of $Y$, refining the cover $\{W_y\}_{y\in\mathcal Y}$. It is easy to check that the cover $\mathcal U_n=\{V\cap f^{-1}(W):V\in\mathcal V,\;W\in\mathcal W\}$ has the properties (1)--(3).

Now let us return to the proof of Theorem 0. On the space $X$ consider the ultrametric $d$ defined by $$d(x,y)=\inf\Big\{\frac1{2^n}:\exists U\in\mathcal U_n\; x,y\in U\Big\}.$$ The condition (1) of Claim guarantees that $d$ is an ultrametric, the condition (2) ensures that the ultrametric $d$ generates the topology of $X$, and the condition (3) guarantees that the fibers of the map $f$ are parallel in the metric $d$.

Remark. The problem posed above has an affirmative answer, given in my next post.

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    $\begingroup$ @user116515 It seems that the second statement in Will Brian's example should be adjusted: one can take a sequence of pairwise disjoint clopen sets of diameter $>1$ in a suitable Cantor set and take other sets be singletons. Then the obtained partition of the Cantor set satisfies $(*)$ but does not admit a "parallel" metric. It seems that in case of a partition one should assume that the partition is both lower semicotinuous (which is the condition (*)) and upper semicontinuous. The absence of the upper semicontinuity yields counterexamples like that of MTyson. $\endgroup$ Oct 27, 2017 at 23:45
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    $\begingroup$ @user116515 On the other hand, if a partition of a space into compact sets is both lower and upper semicontinuous, the the quotient map induced by this partition is perfect and open. So, we fall into the framework of my answer. So, now we should try to exclude the zero-dimensionality, which was essentially used for the construction of the parallel ultrametric. $\endgroup$ Oct 27, 2017 at 23:48
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    $\begingroup$ @user116515 I added to my answer a sketch of the construction of a parallel metric in the general case (of lower and upper semicontinuous partitions). I plan to write a text with a complete proof and post it to Arxiv. $\endgroup$ Oct 28, 2017 at 18:41
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    $\begingroup$ I admit it will take me some time to digest your sketch, especially since it uses terminology with which I am not familiar. But I'm very impressed if you produced some new mathematics from my somewhat idle question. Very nice! $\endgroup$
    – user116515
    Oct 28, 2017 at 19:56
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    $\begingroup$ @user116515 In my next post I wrote a correct (and not very difficult) proof of the general case. $\endgroup$ Oct 29, 2017 at 20:30

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