Let $\ M\ $ be the family of all non-empty bounded regular open subsets of $\ \Bbb R,\ $ where regular means that every $\ G\in M\ $ is equal to the interior of its closure.

Let distance $\ d(G\ H)\ $ be the Hausdorff distance between the closures of $\ G\ $ and $\ H,\ $ for every $\ G\ H\,\in\,M.$

QUESTION: does there exist a function $\ s:\, M\to\Bbb R\ $ that is a metric selection, meaning that:

  • $\forall_{G\in M}\quad s(G)\in G;$

  • $\forall_{G\ H\,\in\,M}\quad |s(G)-s(H)|\ \le\ d(G\ H);$   ?

If yes,

  • can selection $\ s\ $ be injective?


There are a plethora of similar questions. For instance, one may consider metric spaces different from $\ \Bbb R,\ $ e.g. open interval $\ (-1;1)\ $ or perhaps more interestingly, the two-dimensional Euclidean sphere $\ \Bbb S^2,\ $ etc.

It'd be exciting to know how the existence of Hausdorff metric selector depends on the metric space -- say, would the answer be different for different but topologically equivalent metrics of the same topological metrizable space?


One could also ask about Lipschitz selectors (with a fixed constant or arbitrary Lipschitz; or even all continuous, etc.) rather than metric. In particular, condition $\ Lip_2\ $ would provide a much larger family of selectors, when the above metric constrain on selector $\ s\ $ is relaxed to:

$$\forall_{G\ H\,\in\,M}\quad |s(G)-s(H)|\ \le\ 2\cdot d(G\ H).$$

  • $\begingroup$ Why do you include the "regular" hypothesis? $\endgroup$ May 14, 2020 at 4:45
  • $\begingroup$ @EthanDlugie, it's cleaner this way. For instance, the Hausdorff distance between different open sets could be zero. It'd be a nuisance. $\endgroup$
    – Wlod AA
    May 14, 2020 at 4:51
  • 1
    $\begingroup$ I see, e.g. deleting a point from an open interval yields a distinct open set which is not distinct in the Hausdorff metric. Regularity forbids this. $\endgroup$ May 14, 2020 at 4:54
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    $\begingroup$ I guess to this point, no selector exists if you remove the regularity hypothesis. For a selector must assign the same value to sets with Hausdorff distance $0$, but the open sets $G=(-1,1)$ and $H=G-\{s(g)\}$ have Hausdorff distance $0$. $\endgroup$ May 14, 2020 at 5:00
  • $\begingroup$ If you consider the weaker question, without demanding injectivity, then the difference between the two versions (all versus regular) doesn't matter. $\endgroup$
    – Wlod AA
    May 14, 2020 at 5:25

2 Answers 2


No such selector exists. Retooling my comment above, let $G$ be the open interval $(-1,1)$. Take $\epsilon>0$ such that $\ \epsilon<\min(1-s(G), s(G)-1)\ $ hence the closed $\epsilon$-neighborhood of $s(G)$ is contained (comfortably) in $G$. Then $H=G-[s(G)-\epsilon,s(G)+\epsilon]$ is a union of two non-empty open subset of $G$, is regular as per your definition, and $d(G,H) = \epsilon.\ $ But by construction, no element of $H$ lies within distance $\epsilon$ of $s(G)$.

Clearly the problem here is that you're working with open bounded sets, so they don't contain their boundary points. Maybe you can have better luck with compact sets? I thought that was usually the class of subsets to which one applies the Hausdorff metric anyway.

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    $\begingroup$ For compact sets, the $\min$ function works. $\endgroup$ May 14, 2020 at 13:53
  • $\begingroup$ Very good! (at the time of my first reading of your solution, I had my private comprehension problems, sorry). #### <And still, your "clearly" is clearly not clear at all. I mean your general final comment which is false>. $\endgroup$
    – Wlod AA
    May 15, 2020 at 1:10


Theorem  Let $X$ be a metric space containing a homeomorphic copy of the interval $(0, 1)$. Then the regular open sets of $X$ do not admit a uniformly continuous choice function.

I'll show just the case $X = (-2, 2)$ (the interval) and skip the epsilon-delta details and the fact there could be stuff around the embedded path, since the details of this are very similar to the original (see below).

(Note that a choice function admitting an $f$-metric choice function just means uniform continuity from $(S, d_H|_{S \times S})$ to $X$ with $f$ the modulus of continuity.)

For $n \in \mathbb{Z}$ define $$ U_n = (\arctan(n)/\frac{\pi}{2} - \epsilon_n, \arctan(n)/\frac{\pi}{2} + \epsilon_n) $$ where $\epsilon_n$ are sufficiently small so that these sets are disjoint. So we have "order type $\zeta$ many" open intervals side by side inside $(-1,1) \subset X$. Each $U_n$ is a regular open set in $(-1,1) \subset X$, and $U_n \cap U_m = \emptyset$ if $n \neq m$. The union of all these, $U = \bigcup_n U_n$, is also easily seen to be regular open.

Now suppose $g$ is a choice function for regular opens. Then $g(U) \in U_n$ for some $n \in \mathbb{Z}$. Slide $U_L = \bigcup_{m \leq n} U_m$ continuously to the left side of $X$, join it to a single component and morph it into the interval $V_L = (-5/3,-4/3)$. Slide then $U_R = \bigcup_{m > n} U_m$ to the right side, join it to a single component and morph it to $V_R = (4/3,5/3)$. The choice must follow along, i.e. $$ g(U) \in U_L \implies g(V_L \cup V_R) \in V_L. $$

But if we define $U_L' = \bigcup_{m < n} U_m$ and $U_R' = \bigcup_{m \geq n} U_m$, and do the exact same with these sets, we get $$ g(U) \in U_R' \implies g(V_L \cup V_R) \in V_R. $$

That is the contradiction that squares up the proof.


OP has suggested that I write an answer based on my comment. Here's one possible statement you get from that idea, quick write-up, I'll fix later if I screwed up the epsilons.

Let $X$ be a metric space and $S \subset \mathcal{P}(X)$ a set of sets in $X$. Let $f : \mathbb{R}_+ \to \mathbb{R}_+$ be a function. A function $g : S \to X$ is an $f$-metric choice function for $S$ if $g(A) \in A$ for all $A \in S$, and $d(g(A), g(B)) \leq f(d_H(A, B))$ for all $A, B \in S$. We say $S$ then admits an $f$-metric choice function.

Theorem  Let $f : \mathbb{R}_+ \to \mathbb{R}_+$ satisfy $\lim_{x \to 0} f(x) = 0$ and let $X$ be a metric space containing a homeomorphic copy of $S^1$. Then the regular open sets of $X$ do not admit an $f$-metric choice function.

Proof. Let $h : S^1 \to X$ be the embedding of $S^1$ into $X$, and let $\epsilon > 0$ be such that opposite points on $S^1$ map at least distance $\epsilon$ apart form each other in the map $h$. Let $0 < \delta < \epsilon/10$ be such that $f(x) < \epsilon/10$ for $x < 3\delta$.

Identify $S^1$ as $\mathbb{R}/\mathbb{Z}$. To each $a \in S^1$ associate the set $$ k(a) = k_1(a) \cup k_2(a) $$ where $$ k_1(a) = \overline{B_{\delta}(h(a))}^\circ $$ and $$ k_2(a) = \overline{B_{\delta}(h(a+1/2))}^\circ \subset X. $$ If $\delta > 0$ is small enough, $k(a)$ is regular open for all $a$. (The interior of the closure of an open set is regular open, so $k_i(a)$ is. The union of two regular opens may not be regular open in general, but since $\delta < \epsilon/10$ this happens.) Again because $\delta < \epsilon / 10$, the sets $k_1(a)$ and $k_2(a)$ are disjoint.

Suppose we had a choice function $g$ for the regular opens that is $f$-continuous. W.l.o.g. we may assume $g(k(a)) \in k_1(a)$ for some $a \in S^1$. Then by picking small enough increments, it is easy to see that in fact $g(k(a)) \in k_1(a)$ for all $a \in S^1$.

(Here's some algebra to show that in case it's not obvious: If the distance between $h(a)$ and $h(a')$ is at most $\delta$, then the distance between $k_1(a)$ and $k_2(a')$ is at least $\epsilon - 3\delta > \epsilon/10$, and $$ d_H(k(a), k(a')) \leq \max(d_H(k_1(a), k_1(a')), d_H(k_2(a), k_2(a'))) \leq 3\delta, $$ so $g(k(a')) \in k_1(a')$ whenever $g(k(a)) \in k_1(a)$ and $|a'-a|$ is small enough.)

But now we have a contradiction since $$ g(k(a)) \in k_1(a) $$ and $$ g(k(a)) = g(k(a+1/2)) \in k_1(a+1/2) = k_2(a) $$ and $k_1(a) \cap k_2(a) = \emptyset$. Square.

  • $\begingroup$ Your assumption about $\ X\ $ containing a topological circle is painful since it excludes $\ \Bbb R\ $ (and the homeomorphic images of $\ \Bbb R).\ $ Fortunately, you do not need the full force of your assumption. It's enough for your argument to assume that $\ X\ $ contains a non-degenerated arc, a homeomorphic image of $\ [0;1].\ $ Indeed, this will allow you to create a "similar" circle in the space of sets, which circle you can use for your modified proof. ("Similar" with respect to your goal). $\endgroup$
    – Wlod AA
    May 15, 2020 at 1:44
  • $\begingroup$ To be fair, I've mentioned $\ S^2\ $ which contains circles. $\endgroup$
    – Wlod AA
    May 15, 2020 at 1:57
  • $\begingroup$ Of course, the larger a space the harder to find selectors. $\endgroup$
    – Wlod AA
    May 15, 2020 at 2:26
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    $\begingroup$ Not sure how a circle in the space of sets helps. I have a different trick for an arc (for continuous choice) which I can add in a few hours. $\endgroup$
    – Ville Salo
    May 15, 2020 at 6:13
  • $\begingroup$ Ville, you use the circle in $\ X\ $ just to use an induced circle (or closed orbit) in the space of the sets -- that's how your proof works. But when you have that circle in the space of sets (by whichever way), you don't need another one in $\ X.$ $\endgroup$
    – Wlod AA
    May 15, 2020 at 6:40

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