# A question about closed balls in Hilbert Space.

Let $H$ be a separable and infinite dimensional Hilbert Space and let $B$ be a closed ball of $H$ whose diameter is some positive real number. Is every covering of $B$ by closed bounded subsets of $H$, each of which has a diameter less than that of $B$, necessarily infinite? It is easy (some might even say "trivial") to prove that the question has a"yes" answer when the covering sets are closed balls of $H$. But is the answer still "yes" when the covering is by arbitrary closed subsets of $H$ whose diameter is smaller than that of $B$?

• I may be missing something here, but if you have a covering by closed sets with bounded diameter, does this not imply that you have a covering with balls of the same diameter? Commented Mar 6, 2014 at 19:35
• @Thomas An equilateral triangle with side-length $\sqrt{3}$ has diameter $\sqrt{3}$, but the smallest ball (disk) that contains it has diameter $2$. So if you replace a covering set by a ball containing it, you may increase the diameter. Commented Mar 6, 2014 at 20:48
• If M is any closed subset of H whose diameter is d(M) and if D(M) is the diameter of the (unique) smallest closed ball of H that contains M then D(M) is less than or equal to (2^(1/2))*d(M). There exist closed subsets M of H for which d(M) is positive and also D(M) attains this upper bound. One example is the set of all points of H having exactly one non-zero coordinate which is equal to 1. This set has diameter 2^(1/2) and the smallest closed ball of H containing it has diameter 2. Commented Mar 9, 2014 at 20:24

Let $M$ be the diameter of $B$. Each of the closed sets has diameter $< M$, and is therefore contained in a closed ball that also has diameter $<M$. If the covering by arbitrary sets was finite, this would give a covering by finitely many closed balls each of whose diameters was less than $M$.
• A closed set with diameter $d$ need not be contained in a ball with diameter $d$. For example an equilateral triangle with side-length $\sqrt{3}$ has diameter $\sqrt{3}$, but the smallest ball (disk) that contains it has diameter $2$. Commented Mar 6, 2014 at 20:45