Editing Von Neumann bicommutant theorem (section)

===Proof of (i)===
For any {{mvar|x}} and {{mvar|y}} in {{mvar|H}}, the map ''T'' → <''Tx'', ''y''> is continuous in the weak operator topology, by its definition. Therefore, for any fixed operator {{mvar|O}}, so is the map

:<math>T \to \langle (OT - TO)x, y\rangle = \langle Tx, O^*y\rangle - \langle TOx, y\rangle </math>

Let ''S'' be any subset of {{math|''L''(''H'')}}, and ''S''′ its [[commutant]]. For any operator {{mvar|T}} in ''S''′, this function is zero for all ''O'' in ''S''. For any {{mvar|T}} not in ''S''′, it must be nonzero for some ''O'' in ''S'' and some ''x'' and ''y'' in {{mvar|H}}. By its continuity there is an open neighborhood of {{mvar|T}} for the weak operator topology on which it is nonzero, and which therefore is also not in ''S''′. Hence any commutant ''S''′ is [[Closed set|closed]] in the weak operator topology.  In particular, so is {{math|'''M'''′′}}; since it contains {{math|'''M'''}}, it also contains its weak operator closure.