# A peculiar characterization of open balls in a Banach space

Let $E$ be a Banach space and $U$ be a bounded open subset of $E$.

Suppose that for any $x,y\in U$, there exists some open ball $B$ such that $\{x,y\}\subset B\subset U$.

Prove that $U$ is an open ball.

I've been stuck on this seemingly intuitive problem for a day. I've tried contradiction and contraposition, to no avail. I can't even figure why completeness matters here.

I'm looking for a hint that would set me on the right path.

This is a cute problem.

I would look at it like this: if we are going to show $U$ is a ball, we have to identify a good candidate for its radius $R$ and its center $c$.

For $R$, we should think it should be the largest possible radius of a ball contained in $U$. So let $A \subset \mathbb{R}$ be the set of all numbers $r$ such that $U$ contains a ball of radius $r$. Since $U$ is bounded, $A$ is bounded above, so it has a finite supremum. Take $R = \sup A$ to be the supremum.

Now we want to identify the center. Choose a sequence $\{r_n\} \in A$ with $r_n \uparrow R$. Then for each $n$ there is a ball of radius $r_n$ contained in $U$; let $x_n$ be its center, so that $B(x_n, r_n) \subset U$. Intuitively, as $r_n$ gets bigger, the ball $B(x_n, r_n)$ should fill most of $U$, leaving little room to wiggle it around. So it's reasonable to guess that the points $x_n$ are not too far from each other. Indeed, you should try to show that $\{x_n\}$ is Cauchy. By completeness, it converges to some point which we will take as our $c$.

In showing it is Cauchy, you can take advantage of the "two point" property of $U$ in the following way:

Lemma. Suppose $x,y \in E$ and $s,t > 0$ are real numbers. For any $\epsilon > 0$, there exist points $x' \in B(x,s)$ and $y' \in B(y,t)$ such that $\|x' - y'\| \ge s+t+\|x-y\|-\epsilon$.

To see how to choose $x', y'$, draw a picture of $x,y$ and the balls $B(x,s), B(y,t)$. Draw a line through $x$ and $y$.

When trying to show $\{x_n\}$ is Cauchy, you can consider the balls $B(x_n, r_n)$ and $B(x_m, r_m)$. Choosing $x_n', x_m'$ as in the lemma, there is thus a ball containing $x_n', x_m'$ and contained in $U$. Hence its radius is at most $R$. This lets you bound $\|x_n' - x_m'\|$ and therefore also $\|x_n - x_m\|$.

Now that $c,R$ have been defined, you can try to show that $U = B(c,R)$. The $\supseteq$ direction is just a triangle inequality argument. For the $\subseteq$ direction, you'll want to use the fact that $U$ is open, and the lemma and the "two point" property again.