I've seen many references claiming that if $\mathcal{N}$ is a $\sigma$-weakly closed *-subalgebra of a von Neumann algebra $\mathcal{M}$, then by taking $\mathcal{N}_\bot:=\{\phi\in\mathcal{M}_*|a(\phi)=0\text{ for all }a\text{ in }\mathcal{M}\}$ we have $\left(\mathcal{M}_*/\mathcal{N}_\bot\right)^*=\mathcal{N}$ (see e.g. Sakai, 1998). I would like to see a proof of this fact. Any references or inline proofs are much appreciated! I have the vague idea that it has to do with the general result $(V/W)^*=W^\bot$ where $V$ is a vector space, $W$ is a subspace and $W^\bot$ its annihilator.
I am interested in this fact because, since the uniform topology is finer that the $\sigma$-weak one, it proves that $\mathcal{N}$ is a $W^*$-algebra.
Thanks!