# Hausdorff metric selectors

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?

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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?

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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:

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

• Why do you include the "regular" hypothesis? – Ethan Dlugie May 14 at 4:45
• @EthanDlugie, it's cleaner this way. For instance, the Hausdorff distance between different open sets could be zero. It'd be a nuisance. – Wlod AA May 14 at 4:51
• 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. – Ethan Dlugie May 14 at 4:54
• 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 $0$. – Ethan Dlugie May 14 at 5:00
• If you consider the weaker question, without demanding injectivity, then the difference between the two versions (all versus regular) doesn't matter. – Wlod AA May 14 at 5:25

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.

• For compact sets, the $\min$ function works. – Ilya Bogdanov May 14 at 13:53
• 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>. – Wlod AA May 15 at 1:10

edit

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.

original

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.

• 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). – Wlod AA May 15 at 1:44
• To be fair, I've mentioned $\ S^2\$ which contains circles. – Wlod AA May 15 at 1:57
• Of course, the larger a space the harder to find selectors. – Wlod AA May 15 at 2:26
• 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. – Ville Salo May 15 at 6:13
• 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 $\ X.$ – Wlod AA May 15 at 6:40