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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