Connectedness of complement of intersection of two balls

Question

Let $B_r(x_0)\subset \mathcal{R}^n$ be an open ball with radious $r>0$ and centered at $x_0$. Let $y_0\in \partial \overline{B}_r(x_0)$. I want to argue: there exists some $\delta>0$ such that both $B_\delta(y_0)\cap B_r(x_0)$ and $B_\delta(y_0)\setminus B_r(x_0)$ are connected. Is the statement true? If so, how do I make the argument rigorous? If not, could you please provide a counter example?

It is clear that $B_\delta(y_0)\cap B_r(x_0)$ is connected because intersection of two convex sets is convex and hence connected. I am having a hard time proving $B_\delta(y_0)\setminus B_r(x_0)$ is connected. From drawing little diagrams in $\mathcal{R}^2$, my intuition tells me it should be. Any help would be greatly appreciated.

Answer 1

EDIT: this answer was made before the revision of the question.

The answer is probably going to highly depend on the "geometry" of your overarching metric space, that is whether or not there are enough points "around" the set in question so that the complement remains connected.
Think for example of the cross made up of the $x$- and $y$-axes in the real plane $\mathbb{R}^2$, and take $x_0 = 0$, $r = 1$ with the usual Euclidean metric inherited from $\mathbb{R}^2$.
No matter which of the available points $y_0$ and which $\delta > 0$ you'll take, you can imagine that the complement of your desired set cannot be connected.
(I would draw this counterexample but I don't really know how to.)

Answer 2

If we are using the Euclidean metric with $n>1$ there is a quick answer. Choose $\delta> r$. Then $\partial {B}_r(x_0)$ does not intersect $\partial {B}_\delta(y_0)$. The boundary $\partial {B}_\delta(y_0)$ is path connected (see https://math.stackexchange.com/a/1801994/27978 for an idea of how to prove this). Pick $y \in B_\delta(y_0)\setminus B_r(x_0)$. Consider the line $l(t) = x_0+t(y-x_0)$, for $t\ge 1$ $l(t) \notin B_r(x_0)$ and for some $t' \ge 1$ we have $l(t') \in B_\delta(y_0)$. The same process shows that $y_0$ is path connected to $\partial {B}_\delta(y_0)$, hence we have the desired result.

Answer 3

Edit: The following is a proof of the following statement. With notation as in the original question, there exists $\delta' > 0$ such that for all $\delta$ with $0 < \delta \leq \delta'$ the sets $B_{\delta}(y_0)\cap B_r(x_0)$ and $B_r(x_0)\setminus B_\delta(y_0)$ are connected.

This is slightly different from what the question asks for, chiefly in that the question asks for connectivity of $B_\delta(y_0)\setminus B_r(x_0)$. A similar strategy works for that with small enough $\delta$, though another answer provides an argument for large $\delta$.

For convenience, first notice that we can assume that $x_0 = 0$ because the translation $v\mapsto v-x_0$ is a homeomorphism (in fact, an isometry) of all the spaces of interest.

Take $\delta = r/2$ and convince yourself that the resulting complement of interest $C$ is path connected. It suffices to show that there is a path in $C$ starting at any point $x\in C$ and ending at $0$.

Reduce to the 2D case by finding a path contained in a plane $P$ which contains $x_0,x$ and $y_0$. There is an isometric embedding $\pi :P\cap (\overline{B_{r/2}(y_0)} \cup B_r(0)) \to \mathbb{R}^2$ with image $\overline{B_{r/2}(\pi(y_0))} \cup B_r(0) \subseteq\mathbb{R}^2$. You can write this projection down in linear algebra terms directly, if desired, by taking an orthonormal basis for $P$. In that case, $\pi$ is the map given by taking coordinates relative to the basis. The inverse is given by the appropriate matrix multiplication.

The problem has been reduced to finding a path between $0$ and $\pi(x)$ in $B_r(0)\setminus \overline{B_{r/2}(\pi(y_0))} $. But this is straightforward. Note that rotations are isometries in $\mathbb{R}^2$, so we can assume $\pi(y_0)$ is on the vertical axis for convenience. Consider the vertical line centered at $\pi(x)$ and the horizontal line centered at $0$ and convince yourself that they intersect on the interior of $B_r(0)\setminus \overline{B_{r/2}(\pi(y_0))}$, which yields the desired path.

Answer 4

For brevity, I’ll call your points $x_0$, $y_0$ just $x$, $y$, and your balls $B = B_{r}(x)$, $B' = B_{\delta}(y)$. I claim that for every $\delta \in (0,r)$, $B' \setminus B$ is connected. (In fact it’s true for all $\delta > 0$, but I’ll only show it for $\delta < r$.) Here are sketches for a couple of simple and geometrically intuitive arguments:

(1) Let $z$ be the point of $\partial B'$ directly opposite $x$. Then (a) some small ball around $z$ is disjoint from $B$, so its intersection with $B'$ is convex, hence a connected subset of $B'\setminus B$; and (b) for any point in $B'\setminus B$, the line segment from it to $z$ is disjoint from $B$ (since distance from $x$ increases as you move towards $z$), and so gives a path from that point into the known connected subset.

(2) Taking the problem as given for dimension 2, we can get it for higher dimensions as follows: Take $z$ to be any point in $B' \setminus B$ lying on the same diameter of $B'$ as $x$, so that $(x,y,z)$ are collinear — for instance, $z := y - \varepsilon(x - y)$ for some small $\varepsilon > 0$. Now I claim any point $w$ in $B' \setminus B$ has a path to $z$ lying within $B \setminus B'$: Given any such $w$, there’s some plane $P$ containing all four mentioned points (generally, any three points in $\mathbb{R}^n$ lie in some plane; taking a plane containing $x$, $y$, and $w$, then it must also contain $z$ since $(x,y,z)$ are collinear). Now intersecting everything with this plane, we reduce to the 2-dimensional version of this situation, so taking that as known, $w$ must have some path to $z$ lying within $(B \setminus B') \cap P$. This shows that the problem is essentially just 2-dimensional.