Intersection of balls in $l^{\infty}$

Question

Must it be the case that the intersection of any two closed balls in $\ell^{\infty}$ either is empty, a single point, or contains a non-empty open ball?

Answer 1

To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,\ldots )$ and $x_2:=(2,0,0,0,\ldots )$ and then $$ \bar{B_i}(x_i,1):=\{ y\in \ell_\infty : \|y-x\|_\infty \leq 1 \}$$ Since $\|x\|_\infty = \max_{i\in {\mathbb N}} |x_i|$ we see that the point $(1,0,0,0,\ldots )$ will lie in both $\bar{B_1}$ and $\bar{B_2}$. In fact, you should be easily able to show that $$\bar{B_1} \cap \bar{B_2} = \{ (1, y_1, y_2, y_3, \ldots ) : |y_i|\leq 1 \: \forall i \in {\mathbb N} \} $$ So: the intersection is

1. clearly non-empty
2. contains infinitely many points
3. does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.