Does the union of two not disjoint open balls always contain the line connecting the two centers?

Question

Maybe I have been drawing wrong, but the intuition I got from drawing 2d circles suggests the above statement may be correct. I have yet to come up with a proof in $\mathbb{R}^n$ or maybe normed or metric spaces.

Not sure about the most general settings that allows for the above statement to hold, I think I will have to restrict the statement to Hilbert-Spaces because it is easy to draw counter-examples in other norms. The crux seems to be that for two not disjoint open balls $B_{r_1}(x)$ and $B_{r_2}(y)$ in norms other than 2, the line $\overline{xy}$ does not need to cross the intersection $B_{r_1}(x) \cap B_{r_2}(y)$. So I think the statement I want to prove is this:

Let $H$ be a Hilbert space and $x,y \in H$, $r_1, r_2 > 0$. If $B_{r_1}(x) \cap B_{r_2}(y) \neq \emptyset$ then $$\overline{xy} \cap (B_{r_1}(x) \cap B_{r_2}(y)) \neq \emptyset.$$

Does the above statement maybe characterize a certain type of geometric object? I am thinking of a statement like: $A$ a subset s.t. the union of $A$ with a subset $B$ contains a line connecting their "center of mass".

Answer 1

This is true in any metric space whatsoever, if by "the line between $x$ and $y$" you mean "the set of points $z$ such that $d(x,z)+d(y,z)=d(x,y)$". This includes the standard lines in ${\bf R}^n$, as well as normed spaces and geodesic spaces (in non-uniquely geodesic space, e.g. a sphere with the intrinsic metric, this "line" will be the union of geodesics, or at least contain it).

Take any two intersecting balls $B_x=B(x,r_x)$ and $B_y=B(y,r_y)$. Take any point $z$ on the line between $x$ and $y$. Then by definition, $d(x,z)+d(y,z)=d(x,y)$. But since the two balls intersect, $r_x+r_y>d(x,y)$, and hence $d(x,z)+d(y,z)<r_x+r_y$, so either $d(x,z)<r_x$ or $d(y,z)<r_y$, and hence $z\in B_x$ or $z\in B_y$.

Answer 2

If the intersection of the open balls $B(x_1, r_1)$ and $B(x_2, r_2)$ is non empty then $$\|x_1-x_2\|\leq \|x_1-y\|+\|y-x_2\|<r_1+r_2$$ where $y$ is a point of the intersection.

The points on the segment $\overline{x_1 x_2}$ which connects the centers $x_1$ and $x_2$ are given by $P(t):=tx_1+(1-t)x_2$ with $t\in[0,1]$.

Then for $t\in [r_2/(r_1+r_2),1]$: $$\|P(t)-x_1\|=(1-t)\|x_2-x_1\|<(1-t)(r_1+r_2)\leq r_1,$$ that is $P(t)$ belongs to $B(x_1, r_1)$.

In a similar way for $t\in [0,r_2/(r_1+r_2)]$: $$\|P(t)-x_2\|=t\|x_2-x_1\|<t(r_1+r_2)\leq r_2,$$ that is $P(t)$ belongs to $B(x_2, r_2)$.

Hence $P(t)$ is in the union of the two balls for any $t\in[0,1]$.