Ramana Kumar comments on Eight Definitions of Observability

Ramana Kumar 29 Jan 2021 8:26 UTC
LW: 3 AF: 2
AF
Because $S_{1}$ and $S_{2}$ are not a partition of the world here.
EDIT: but what we actually need in the proof is ${A s s u m e}_{S_{1}} (C) & {A s s u m e}_{S_{2}} (C) & \dots ≃ {A s s u m e}_{S_{1} \cup S_{2}} (C) & \dots$ where the $\dots$ do result in a partition, so I think this will work out the same as the other comment. I’m still not convinced about biextensional equivalence between the frames without the rest of the product.
- Ramana Kumar 29 Jan 2021 10:52 UTC
  LW: 3 AF: 2
  AF Parent
  And it seems we do actually need ${A s s u m e}_{S_{1}} (C) & {A s s u m e}_{S_{2}} (C) ≃ {A s s u m e}_{S_{1} \cup S_{2}} (C)$ in the proof to justify:
  Thus it suffices to show that $(D_{1} & 1_{R_{2} \cup R_{3}}) \otimes (D_{2} & 1_{R_{1} \cup R_{3}}) ≃ D_{1} & D_{2} & 1_{R_{3}}$ .
  Without it, we have to show $(D_{1} & 1_{R_{2} \cup R_{3}}) \otimes (D_{2} & 1_{R_{1} \cup R_{3}}) ≃ {A s s u m e}_{S_{1} \cup S_{2}} (C) & 1_{R_{3}}$ instead.
  - Ramana Kumar 30 Jan 2021 0:12 UTC
    LW: 3 AF: 2
    AF Parent
    UPDATE: I was able to prove ${A s s u m e}_{S_{1}} (C) & {A s s u m e}_{S_{2}} (C) ≃ {A s s u m e}_{S_{1} \cup S_{2}} (C)$ in general whenever $S_{1}$ and $S_{2}$ are disjoint and both in $O b s (C)$ , with help from Rohin Shah, following the “restrict attention to world $S_{1} \cup S_{2}$ ” approach I hinted at earlier.