# Norm bounded sets in $C^*$-algebras/von Neumann algebras

I was reading about the weak operator topology in wiki, and I saw that: Norm-bounded sets in $$B(H)$$ are pre-compact in WOT.

I was wondering: Is there an analogous statement for von Neumann algebras? $$C^*$$-algebras?

## 1 Answer

The statement would have been that norm bounded sets are pre-compact, as being bounded does not guarantee that it is closed.

Other than that, yes. You would usually see a von Neumann algebra $$M$$ as embedded in $$B(H)$$. So a wot-closed bounded set is a closed subset of a compact set (a closed ball in $$B(H)$$) and thus compact.

In the case of a C$$^*$$-algebra, you would have to embed it in some $$B(H)$$ to even have a wot topolgy. In that case, bounded sets would be pre-compact.

• By pre-compact, you mean has compact closure? Mar 10, 2021 at 19:08
• Yes. $\ \ \ \$ Mar 10, 2021 at 19:09
• How about the following? Suppose $S\subset M$ is norm-bounded. Since $M=(M_*)^*$ (an isometric isomorphism), the unit ball in $M$ is $\sigma(M,M_*)$-compact (not sure if I'm saying it right). Then $S$ has $\sigma(M,M_*)$-compact closure. Mar 10, 2021 at 19:21
• Yes. That's the classic Banach-Alaoglu Theorem. Mar 10, 2021 at 19:40
• $M$ is a von Neumann algebra in the argument I gave. In the argument you gave, the WOT for $M$ is induced by the containment $M\subseteq B(H)$. You get a non-equivalent topology if you use a different $B(K)$, no? Mar 10, 2021 at 19:45