Editing Von Neumann bicommutant theorem (section)

===Proof of (ii)===
This follows directly from the weak operator topology being coarser than the strong operator topology: for every point {{mvar|x}} in {{math|cl<sub>''S''</sub>('''M''')}}, every open neighborhood of {{mvar|x}} in the weak operator topology is also open in the strong operator topology and therefore contains a member of {{math|'''M'''}}; therefore {{mvar|x}} is also a member of {{math|cl<sub>''W''</sub>('''M''')}}.