Connectedness of balls in a compact, connected metric space

Let $$(X,d)$$ be a compact, connected metric space. For every $$\epsilon>0$$ define an equivalence relation on $$X$$ by $$x\sim_{\epsilon}y$$ if and only if there exists a finite sequence $$(x=x_0,x_1,\dots,x_n=y)$$ such that $$d(x_i,x_{i+1})<\epsilon$$.

Note that the space is connected if and only if for every $$\epsilon>0$$, the $$\epsilon$$-equivalence class of every point is the whole space. See this answer. My interest in this collection of equivalence relations is their properties when one restrict them to certain subsets of the space: For a set $$A \subset X$$ and $$\epsilon>0$$, define $$a \sim_{\epsilon}^{A} b$$ if and only if there exists a $$\epsilon$$-step sequence between $$a$$ and $$b$$ contained in $$A$$.

The following property is something intuitive one could expect to hold in any compact, connected metric space:

Let $$U$$ and $$V$$ be open disjoint subsets of $$X$$ and denote $$K:=(U \cup V)^\complement$$. Let $$\epsilon>0$$ and let $$u \in U$$ and $$v \in V$$ with $$u\sim_{\epsilon}^{U \cup V} v$$ through a finite sequence $$S_{\epsilon}(u,v) = (u=x_0,x_1,\dots,x_n=v) \subset U \cup V$$. Then there exists some $$w \in S_{\epsilon}(u,v)$$ with $$d(w,K)<\epsilon$$.

The proof I managed to find consists on the additional assumption that every ball is a connected subset of the space. Assuming this, one can easily see that there exists some ball $$B$$ of radius $$\epsilon$$ intersecting $$U$$ and $$V$$, so assuming the ball is connected the proof is almost immediate.

I couldn't find a counterexample to this property for compact, connected metric spaces that contain some disconnected ball. Yet, I couldn't prove it when I removed the assumption.

• I can find some $w$ and $w'$ in $\bar U\cap \bar V$ such that $\forall e>0\,(w\sim_e u \land w'\sim_e v)$ but I have not shown that $w$ could equal $w'$.... (Note that $\bar U\cap \bar V$ is disjoint from $U\cup V$ .) – DanielWainfleet Jul 23 at 3:29
• I believe I do not understand the property you are considering correctly. If I take any $w \in (U \cup V)^c$ then, since $X$ is connected, I have $u \sim_\varepsilon w$ and $w \sim_\varepsilon v$ for every $u \in U$ and $v \in V$ without any condition on $u$ and $v$. What am I missing? – Matthias Klupsch Jul 23 at 6:10
• You are definitely right, I should rewrite it as follows: There can be found w such that its distance from the given sequence inside U\capV is at most epsilon. – NAR Jul 23 at 6:48
• Edit your question again: The highlighted sentence is currently meaningless. – Moishe Kohan Jul 23 at 7:01
• Are you sure this is the right question? The point 𝑣 plays essentially no role. As it stands, your question has easy positive answer. (Just take any $w$ on the boundary of the connected component of $U$ containing $u$.) – Moishe Kohan Jul 23 at 7:36

Consider the unit circle $$S^1$$ with its standard angular distance function $$d$$. For non-antipodal points $$p, q\in S^1$$ let $$pq$$ the closed arc with the end-points $$p, q\in S^1$$ and of the length $$<\pi$$. Now, for $$\epsilon$$ satisfying $$0<\epsilon<\pi/2$$, let $$uv, ab$$ be two such arcs of the lengths $$\epsilon/2$$ and $$\epsilon/4$$ respectively such that $$ab$$ is contained in the interior of $$uv$$. Let $$X$$ denote the metric space obtained by removing the arc $$ab$$ from $$S^1$$ (and keeping the distance function). Then $$X$$ is clearly connected. Let $$k\in S^1$$ be the point antipodal to the midpoint of the arc $$ab$$. Let $$U, V$$ denote the connected components of $$X -\{k\}$$ containing $$u, v$$ respectively. Thus, $$K=\{k\}$$ is the complement to $$U\cup V$$ in $$X$$ and $$U\cap V=\emptyset$$, while $$U, V$$ are both open in $$X$$. At the same time, $$d(u,k)\ge \pi- \epsilon > \epsilon$$ and the same for $$v$$. Thus, $$K$$ contains no points within distance $$\epsilon$$ from the $$\epsilon$$-chain $$\{u, v\}$$.
• Thanks for the detailed example. Maybe I'm missing something basic, but it seems to me that removing the closed arc $ab$ from $S^1$ makes $X$ not compact. – NAR Jul 24 at 20:36
• As for the last comment in your answer, maybe I should mention my motivation. I basically want to prove the following: Let $x_n$ be a sequence in a compact, connected metric space, with the property $d(x_n,x_{n+1})$ converges to $0$. Then the partial limits set of $x_n$ is a connected set. It is quite simple to prove it using the property mentioned in my question. – NAR Jul 24 at 20:40
• @NAR: Just remove the open arc $ab$ instead. – Eric Wofsey Jul 24 at 21:54
• Sure, thanks. It indeed works, but the property still holds for every $\epsilon$ sufficiently small, so it is not a "full" counterexample to me. – NAR Jul 25 at 1:53
• @NAR: I suggest you think more about your question and about all the quantifiers missing there. In your question you said is "let $\epsilon$". I interpreted this as asking for the property to hold for all $\epsilon>0$, so a counter-example would mean to find some $\epsilon>0$. This is what I found. Needless to say, I still do not know what your question really is. – Moishe Kohan Jul 25 at 2:04