# Is the ideal of compact operators strongly Borel?

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 $\mathcal{B}(H)$?

Remarks:

• 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.

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Yes.

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 $\mathcal{B}(H)$.

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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