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Oct 28, 2022 at 1:56 comment added Nik Weaver I understand. Glad that Matthew Daws's excellent answer helped!
Oct 28, 2022 at 0:18 history edited Sean CC BY-SA 4.0
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Oct 28, 2022 at 0:17 comment added Sean Thanks @NikWeaver, you are indeed correct and this is how my proof went also. It was not communicated very well in my original post, but my main confusion was regarding the existence of support projections in the first place. I had come with intuition from measure theory rather than vnAs, which led me astray.
Oct 28, 2022 at 0:10 vote accept Sean
Oct 27, 2022 at 11:08 comment added Nik Weaver The closed support projection is surely the meet of the set $\{p \in A'': p$ is closed and $\phi(p) =1\}$. Everything in this set dominates the usual support projection, and every closed projection that dominates the support projection is in this set, so yes, the closed support projection is the closure of the ordinary support projection.
Oct 27, 2022 at 9:27 history edited Matthew Daws
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Oct 27, 2022 at 9:26 answer added Matthew Daws timeline score: 4
Oct 25, 2022 at 3:14 history asked Sean CC BY-SA 4.0