MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $H$ be a separable infinite dimensional Hilbert space. Denote by $\mathcal{B}(H)$ the space of bounded operators on $H$, and $\mathcal{K}(H)$ the ideal of compact operators. When endowed with the strong (or weak) operator topology, is $\mathcal{K}(H)$ Borel in $\mathcal{B}(H)$?


  • It is well known that, unlike in the norm topology, $\mathcal{K}(H)$ is not closed; the identity operator is a strong limit of finite rank projections.

  • $\mathcal{K}(H)$ is analytic: $\mathcal{K}(H)$ is Polish in the norm topology, and the inclusion map $\mathcal{K}(H)\to\mathcal{B}(H)$ is norm-weak continuous.

share|cite|improve this question
up vote 4 down vote accepted


This can be deduced from the argument given for Corollary 3.2 in G. A. Edgar, Measurability in a Banach space, Indiana Univ. Math. J. 26 (1977), 663-677, MR542944.

The proof (attributed to Talagrand) is easy, so I reproduce it:

Choose a countable norm-dense subset $\lbrace d_k : k \in \mathbb{N}\rbrace \subseteq \mathcal{K}(H)$ and let $B$ be the unit ball in $\mathcal{B}(H)$. Then $B$ is compact in the weak operator topology and $$ \begin{align*} \mathcal{K}(H) & = \bigcap_{n \in \mathbb{N}} \left(\mathcal{K}(H) + \tfrac{1}{n}B\right) \supseteq \bigcap_{n\in\mathbb{N}}\bigcup_{k \in \mathbb{N}}\left(d_k + \tfrac{1}{n}B\right) \supseteq \mathcal{K}(H) \end{align*} $$ shows that $\mathcal{K}(H)$ is a $K_{\sigma\delta}$ in $\mathcal{B}(H)$ with the weak operator topology.

Therefore $\mathcal{K}(H)$ is Borel with respect to all the usual topologies considered on $\mathcal{B}(H)$.

share|cite|improve this answer
The fact that $\mathcal{K}(H)$ is Borel in the weak operator topology is a special case of Edgar's Theorem 3.1. The same argument shows that every norm-separable subalgebra of $\mathcal{B}(H)$ is Borel in the weak operator topology. – Martin Jun 20 '13 at 9:00
Thanks for the answer, and the link to this paper! – Iian Smythe Jun 20 '13 at 13:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.