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.


1 Answer 1



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 $\mathcal{B}(H)$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Martin
    Commented Jun 20, 2013 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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