It is well-known that any finite-dimensional normed space is isomorphic to , where . Therefore, the Heine-Borel theorem tells us that closed and bounded subsets of are compact. We can say that has the Heine-Borel property. Perhaps less well-known to beginning students is that an infinite-dimensional normed space need not have the Heine-Borel property. Taking our cue from Big Rudin Chapter 4 Exercise 7, we prove in this post that no infinite-dimensional Hilbert space has the Heine-Borel property.
Let be an infinite-dimensional Hilbert space, and let be an orthonormal system. Then is closed and bounded but not compact (thus, disproving the Heine-Borel theorem in infinite-dimensions).
Proof. Clearly, is bounded by the constant . To see that is closed, suppose is some sequence in such that . I claim that for all sufficiently large. Otherwise, it follows from Bessel’s inequality that
which is a contradiction. This proves that . To see that is not compact, observe that the sequence is not Cauchy since by orthonormality,
so that contains no convergent subsequences.
Let be a sequence of positive numbers, and let be the closure of the set
is compact if and only if .
Proof. If is compact, then is bounded and therefore there exists some positive constant such that
which implies that .
Now suppose that . Let be some sequence in . I claim that , for each . Indeed, fix . By definition of , there exists a finite linear combination such that and . By the triangle inequality and Cauchy-Schwarz,
Since was arbitrary, we see that , , and is a bounded sequence in , for each . By the Bolzano-Weierstrass theorem, there exists a covergent subsequence . By induction, we can construct nested subsequences such that
converges to limit , for each
Set . Since , the series converges in . I claim that the sequence converges to . Indeed, by uniform convergence we can evaluate the limit term-by-term to obtain
where we use the continuity of the inner product to obtain the penultimate equality.
Lastly, we remark that is not a locally compact space by Riesz’s lemma.