Lemma 37.2.5.label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, $T \in A''$, and $x \in H$, then $Tx \in \ol{\bracsn{Sx|S \in A}}$.
Proof, [Lemma 18.4, Zhu93]. Let $M = \ol{\bracsn{Sx|S \in A}}$, then since $A$ is self-adjoint, $M$ is a reducing subspace for each element of $A$. Let $P \in B(H)$ be the orthogonal projection onto $M$, then $PS = SP$ for all $S \in A$. As such, $P \in A'$.
Now, since $T \in A''$, $TP = PT$ as well, so $M$ is a reducing subspace for $T$. As $A$ is unital, $x \in M$, so $Tx \in M = \ol{\bracsn{Sx|S \in A}}$.$\square$
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