# Prove the unit ball in the space $\ell_2$ of infinite sequences is not totally bounded.

Prove that the unit ball $B$ in the space $\ell_2$ of infinite sequences $\{x_n\}$ satisfying the condition: $\sum^\infty_{n=1}x^2_n \leq 1$ is not totally bounded.

To begin, I would like to say I know the concept of what a totally bounded metric space $M$ means. I can see that the set of all points $x=(x_1,x_2,\ldots,x_n,\ldots)$ satisfying the above condition is a convex body (a non empty convex set with a non empty interior). It follows that the interior is the set of all points $x=(x_1,x_2,\ldots,x_n,\ldots)$ satisfying $\sum^\infty_{n=1}x^2_n < 1$. But how am I to show that the ball is not totally bounded in a formal proof? Help/hints greatly appreciated, thanks.

Consider the set $\{v_n : n=1,2,3,\ldots\}$ in $\ell_2$ where $v_n$ is the point $(x_1,x_2,x_3,\ldots)$ for which $x_n=1$ and $x_k=0$ for all $k\ne n.$ What happens if you try to cover it with balls of radius $1/10\text{?}$ Each ball can cover only one of these points. All of these points are on the boundary of the closed unit ball. Thus the closed unit ball cannot be covered by only finitely many open balls of radius $1/10.$
• Interesting, thanks! So you considered a set of sequences such that it was a set of discontinuous sequences in the sense that sequences are isolated from each other? And all of these isolated points are on the boundary of the closed unit ball because you made $x_n=1$? The one thing I do not follow is why each ball can cover only one of these points, sorry if I am missing something trivial. – Rick Owens Apr 14 '17 at 2:10
• Must you first define the metric of the space $\ell _2$? – Rick Owens Apr 14 '17 at 2:12
• The metric is the one following from the norm -- you have the norm $\ell_2$, so the metric follows from that: $d(x,y) = \lVert x-y\rVert_2$ – Clement C. Apr 14 '17 at 2:13
• @ReginaldDick : We have, for example, $$v_3 = (0,0,1,0,0,0,0,\ldots)$$ where all entries are $0$ except the third one, and $$v_3 = (0,0,0,1,0,0,0,0,\ldots,$$ etc. The norm is $\|(x_1,x_2,x_3,\ldots)\| = \sqrt{x_1^2+x_2^2+x_3^2+\cdots}. \qquad$ – Michael Hardy Apr 14 '17 at 2:25