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

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".

• Hint: Connectedness. Commented Aug 20, 2016 at 11:15
• The word "contain" in the title does not seem to match the question. And it would be unlikely anyway, the line through the two centers is unbounded so couldn't be contained in the intersection of two compact objects in $\Bbb{R}^n$ (in the Euclidean metric). Even if you only mean "contains the line segment connecting the two centers", there is no real challenge drawing counterexamples: arrange the two balls a little farther apart so neither includes the other's center. Perhaps "meets" or "intersects" would be words that better capture your intent. Commented Aug 20, 2016 at 19:28

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 $z\in B_y$.

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 $t\in[0,1]$.