## The unit ball of Hilbert space is not compact (p18 (B))

Show that the unit ball $\mathcal{B} := \{x \in \ell^2: \| x \| \le 1 \}$ of Hilbert space is not compact.

Proof.
Let $m \in \mathbf{N}$.  For each $z \in \mathbf{C}, | z | = 1$, define $x_m^z : \mathbf{N} \to \mathbf{C}$ by $x_m^z(n) = 0$, if $n \neq m$, and $x_m^z(m) = z$.  Let $A_m^z := \{ x \in \ell^2 : \| x - x_m^z \| < 1/4 \}$.

Now, let $\mathcal{B}_\delta := \{ x \in \ell^2 : \| x \| = 1 \}$.  Then $\mathcal{B}_\delta = \mathcal{B}_\delta^1 \cup \mathcal{B}_\delta^2$, where $\mathcal{B}_\delta^1 := \{ x \in \ell^2 : \| x \| = 1 \text{ and } \sup_{n \in \mathbf{N}} |x(n)| < 1 \}$, and $\mathcal{B}_\delta^2 := \{ x \in \ell^2 : \| x \| = 1 \text{ and } \sup_{n \in \mathbf{N}} |x(n)| = 1 \} = \bigcup_{m,z} \{ x_m^z \}$.

Now, for each $x \in \mathcal{B}_\delta^1$, $\sup_{n \in \mathbf{N}} |x(n)| = 1 - \epsilon$ for some $\epsilon > 0$.  Then let $A^x := \{ y \in \ell^2 : \| y - x \| < \epsilon \}$.  Finally, let $A^o := \{ x \in \ell^2 : \| x \| < 1 \}$,

Then $\{ A^x, A^z_m, A^o \}$ covers $\mathcal{B}$. Note that for $m \neq n$, $A^{z_1}_m \cap A^{z_2}_n = \emptyset$.  Also, $x_m \notin A^x$ for any $x$, and $x_m \notin A^o$.  Hence for each $m$ we need at least one set from $A^z_m$, for some $z$.  Hence there is no finite subcover.

Oh, I see. $(e_n)$, where $e_n(i) = 1$ if $i = n$ and $0$ otherwise. Actually, my proof above was based on that sequence originally, but I didn’t think about the convergent subsequence characterization of compactness. Oops, lol.