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As lemma 6 on p.44 of Dixmier's book on Von Neumann algebras, he states that if $A$ is a *-algebra (i.e. possibly without identity, not necessarily closed in any topology) of operators in $B(H)$ such that the closed subspace of $H$ generated by the images of the elements of $A$ is actually all of $H$, then $A'' \subset \bar A$ where the closure is in the ultraweak topology. I was hoping someone would help me to prove this, because I believe the book's proof is wrong. What he ends up doing is showing that for every $S \in A''$, $\epsilon>0$ and every $f \in V$ where $V$ is the set of seminorms that generates the ultrastrong topology, there exists $T$ so that $f(T-S)< \epsilon$. But this is insufficient. I need there to be an entire net of $T$s that depends only on $f$. Can someone either give a new argument or explain why the given argument is sufficient? Thanks.

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Well, the set $V$ of seminorms is directed, right? So there's your net. –  t.b. Jul 18 '12 at 4:22
    
It's indexed by the elements of the countable direct sum of H. I don't see how these form a net. Also, I can't say I understand what construction of a net you are suggesting here. This isn't like sequences, where the analogous fact let's me "choose one for $\epsilon$, then $\epsilon/2$ etc." EDIT: Those are two separate concerns btw. –  Jeff Jul 18 '12 at 5:12
    
There are many ways to get a net out of this argument. The way I was suggesting is: The set $V$ is partially ordered by $f \leq g$ if $f(T) \leq g(T)$ for all $T$. If $f,g \in V$ then $h(T) = f(T) + g(T)$ is in $V$ and dominates both $f$ and $g$. For each $f \in V$ pick $T_f$ such that $f(T_f - S) \lt 1$. Then $(T_f)_{f \in V}$ is a net converging to $S$ (since $f \in V$ implies $f/\varepsilon \in V$), so it suffices to show that for every $f$ and every $\varepsilon$ you can find $T$ such that $f(T - S) \lt \varepsilon$ which seems to be what Dixmier is doing. –  t.b. Jul 18 '12 at 7:29
    
So yes, in a way this is exactly like sequences... See also here. –  t.b. Jul 18 '12 at 7:30
    
Thank you, this answers my question. –  Jeff Jul 19 '12 at 0:31

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