## Geometry of the Hilbert sphere

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Let $X$ be the unit sphere in $\ell^2$, i.e. $X=\{x\in\ell^2: \|x\|=1\}$. Let the metric on $X$ be the geodesic metric, i.e. $d(x,y)=\cos^{-1}\langle x,y\rangle$. Call a set a ball-intersection if it is an intersection of closed balls with centers in $X$.

Does there exist a decreasing sequence of nonempty ball-intersections in $X$ with void intersection?

If we let $S_0=\cap_i B[e_i,\pi/2]=\{x\in X: x_i\ge 0 \forall i \},$ $S_1=\{x\in S_0: x_1=0\},$ $S_2=\{x\in S_0: x_1=x_2=0\}, \cdots$, then $\cap_i S_i=\emptyset$. But $S_i$ for $i\neq 0$ are not ball-intersections.

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$S_n=S_0\cap B[-e_1,\pi/2]\cap\dots\cap B[-e_n,\pi/2]$, so it is a ball-intersection.  – Anton Petrunin Nov 2 2011 at 1:41
@Anton. Thanks for pointing this out. – TCL Nov 2 2011 at 2:05

## 1 Answer

I think yes. Your balls are of the form $X\cap S(x,r)$ where $x\in X$ and $S(s,r)$ is the slice `$\{y: \|y\|\le 1 \ \text{and} \ \langle x,y\rangle \ge r\}$` of the unit ball. Note that if $y$ is in a slice, so is $y/\|y\|$. The slices have non empty intersection because they are weakly compact.

EDIT: This argument looks OK if all $r$ are positive, but $r$ can be negative, so some further thought is needed.

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@Bill. Fortunately, for my purpose $r\ge 0$ is what I need. Thank you. – TCL Nov 2 2011 at 1:35
@TCL, it works only for $r>0$. – Anton Petrunin Nov 2 2011 at 1:49
@Anton. You are right. $r>0$ is what I need. – TCL Nov 2 2011 at 2:03
Good, then I won't think about the general case. – Bill Johnson Nov 2 2011 at 2:46
The idea works for closed geodesic convex subsets of $X$ with diameter  $\le k<\pi/2$. – TCL Nov 2 2011 at 11:17
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