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.

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This entry was posted in Descriptive Set Theory, Kechris, Logic. Bookmark the permalink.

2 Responses to The unit ball of Hilbert space is not compact (p18 (B))

  1. Jindrich Zapletal says:

    This is way too involved, and I do not quite get it. Is it not easier to find a (completely obvious) sequence without any converging subsequence?

  2. alanmath says:

    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.

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