Compact Sets and Compact Operators

Proof. Properties 2 and 3 are left to the reader. For property 1, assume that S is an unbounded compact set. Since S is unbounded, we may select a sequence {vn}n=1 such that ‖vn‖ → 0 as n→∞. Since S is compact, this sequence will have a convergent subsequence, say {vk}k=1, which will still be unbounded. This sequence is Cauchy, so there is a positive integer K for which ‖v`− vm‖ ≤ 1/2 for all `,m ≥ K. Fix ` and note that by the triangle inequality ‖vm‖ ≤ 1/2 + ‖v`‖. Now, the right side is bounded, because ` is fixed. However, ‖vm‖ → ∞ as m → ∞. This is a contradiction, so S must be bounded. For property 4, let S = {f ∈ H : ‖f‖ ≤ 1}. Every o.n. basis {φn}n=1 is in S. However, for such a basis ‖φm−φn‖ = √ 2, n 6= m. Again, this means there are no Cauchy subsequences in {φn}n=1, and consequently, no convergent subsequences. Thus, S is not compact.