An example of a closed bounded subset of a Hilbert space that is not totally bounded

Question

Let $l^2$ be the set of complex square summable series. Can you give an example of a subset of $l^2$ which is bounded and closed but not totally bounded. I’m new to this concept so I’m not that good at coming up with examples.

Answer 1

The closed unit ball $B=\{x:\|x\|\le 1\}$ is one such example. That its complement is open can be shown using the triangle inequality (this part is not specific to Hilbert space; this kind of reasoning works in any metric space.) The lack of total boundedness follows from the fact that $B$ contains all standard basis vectors $$(1, 0, 0, 0, \dots), \ (0, 1, 0, 0, \dots), \dots$$ The distance between any two of such vectors is $\sqrt{2}$. If $B$ is covered by finitely many set, some of these sets must contain more than one of the basis vectors, and therefore have diameter at least $\sqrt{2}$.