Ultra weakly closed *-subalgebra of B(H) I'm currently working on a text about von Neumann algebras and the author used without further clarifying that any ultra weakly closed *-subalgebra of $B(H)$ contains a largest projection. Could someone tell me why? 
 A: Let
$$
V := AH^{\perp} = \{x \in H : x\perp ay \quad\forall a\in A, y\in H\}
$$
Let $W$ be the orthogonal complement of $V$, then $W$ is a non-degenerate representation of $A$. It is a consequence of the Double Commutant theorem that
$$
\text{id}_W : W\to W
$$
is an ultra-strong limit of elements of $A$. Hence, $A$ contains $\text{id}_W$, which can be identified with the orthogonal projection $e\in B(H)$ onto $W$. Furthermore,
$$
A = Ae
$$
So for any other projection $f\in A = Ae$, we must have
$$
f = fe
$$
and so $f\leq e$. Hence, $e$ is the largest projection in $A$.
A: Say $M\subset B(H)$ is your von Neumann algebra. Let $\mathcal P=\{p\in M:\ p \text{ is a projection }\}$.  Let $K\subset H$ be the subspace generated by $\bigcup_{p\in\mathcal P}pH$, and write $p_K$ for the orthogonal projection onto $K$. 
Then for any $q\in \mathcal P$, $p_Kq=q$. This implies that $q\leq p_K$, and also that $p_kq=qp_K$. For any projection $r\in M'$, 
$$
rpH=prH\subset pH,
$$
so $rK\subset K$. This implies that $p_Krp_K=rp_K$, from where it follows that $rp_K=p_Kr$. So $p_K\in M''=M$. 
The above is a sketch of the proof that in a von Neumann algebra arbitrary unions of projections are again in the algebra. 
