1

Let $H$ be a infinite dimensional Hilbert space and $B(H)$ be the space of bounded and linear operators on $H$. I know that weak operator topology (wot) and strong operator topology (sot) are metrisable on bounded subsets of $B(H)$. I think to show wot is not metrisable on $B(H)$, I should show $B(H)$ is not first countable. But I do not have any idea to do it. Please help me. Thanks in advance.

niki
  • 3,372
  • 1
  • 15
  • 35
  • The proof of this is non trivial. For a hint, see http://math.stackexchange.com/questions/424876/weak-topology-on-an-infinite-dimensional-normed-vector-space-is-not-metrizable – Crostul Apr 18 '15 at 22:57
  • @Crostul: But your hint is about weak topology, and my question is about weak operator topology. unfortunately, I can not know how to use it. – niki Apr 19 '15 at 07:19
  • HINT: $B(H)$ is a Hilbert space. – Crostul Apr 19 '15 at 09:19

0 Answers0