# Some questions about “inspecting” the boundary of a closed ball in Hilbert space

Let H be a separable Hilbert space and suppose that H is infinite dimensional. Let B be a closed ball of H-which has a positive radius-and let S be the boundary of B. A non-empty subset C of H is an "inspection set for S" just in case (1) C and B are disjoint and (2) If p is any point of S, there exists a point q of C such that the straight line segment whose end points are p and q, contains no interior point of B. In other words, it should be possible to "see" each point of S from at least one point of C. ...... My questions are about the kind of subset C of H that can be an "inspection set for S". Can C be a simple closed curve? Can C be compact? Can C be a homeomorphic image of a sraight line? I suspect that the answer to all these questions is negative and not difficult to prove (although I have gotten nowhere trying to prove it myself). The answers are, of course, clearly positive when H is finite dimensional, since we can then consider H to be a finite dimensional Euclidean space.

• An idea: Assume $B$ is centered at the origin and consider a two-dimensional subspace $P\subset H$. If $C$ is a spiral on $P$ (it has points arbitrarily far in every direction), then the only points of $S$ that are not visible from $C$ are those which are orthogonal to the plane $P$. Now if you add suitable wiggles to $C$ in directions orthogonal to $P$, you might be able to see all of $S$. If this construction works in the infinite dimensional setting (it works in Euclidean spaces), you could get $C$ homeomorphic to a line. – Joonas Ilmavirta Jan 31 '15 at 21:52

## 1 Answer

(I'll do this for real Hilbert spaces, which seems natural here.) Just as in finitely many dimensions, a point $x\in H$ sees exactly the subset $$A(x)=\{ s\in S: \langle x, s\rangle \ge 1\} .$$ Since we cannot cover $S$ by finitely many sets of the type $\bigcup A(x)$, with the union taken over $x$ from a small ball, say, it follows that $C$ can not be compact.

On the other hand, covering $S$ with countably many sets $A(x_j)$ is no problem, so $C$ can be homeomorphic to a line. More explicitly, fix $x_1,x_2,\ldots$ such that every point on $S$ is seen by an $x_j$. We can also insist that $\|x_{n+1}\|\ge \|x_n\|+1$, say. Then define a bijective continuous function $f:\mathbb R\to H$ that satisfies $f(n)=x_n$.

• Your nice covering certainly shows that C can be the homeomorphic image of a straight line-which I had thought to be impossible. In fact C can be the homeomorphic image of a ray. You also point out that C must be an infinite set and can be countable. But how, then, to show that C cannot be an infinite compact set, does not seem so obvious-although I think I can see a way. – Garabed Gulbenkian Feb 2 '15 at 19:46
• @GarabedGulbenkian: If $C$ were compact, then I could cover $C$ by finitely many small balls, but this would lead to the scenario that I refute in the first paragraph ("Since we cannot cover $S$ ..."). – Christian Remling Feb 3 '15 at 1:21
• I see what you are saying. A compact subset of any metric space M, is always a subset of a finite union of arbitrarily small balls of M. If H is the metric space and U is any finite union of sufficiently small balls of H, then it is easy to see that there exist points of S which are not "visible" from any point of U. I was confused by the (irrelevant) fact that-since H is infinite dimensional-no ball of H, however small, can be compact. – Garabed Gulbenkian Feb 3 '15 at 20:23