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Mar 12, 2023 at 10:16 comment added Monroe Eskew I don’t see any way to show it in general besides noting that generic filters for nontrivial forcings don’t exist (in the ground model).
Mar 12, 2023 at 10:06 vote accept Connor W
Mar 12, 2023 at 10:06 comment added Connor W Thanks, Monroe. Is there a more direct way to see that a strong master condition can't exist for a definable class embedding? I don't mean directly in the precise sense, I mean an argument which refers to the actual embedding and poset, rather than a "Oh, if there were a strong master condition, then you could build an $M$-generic internally." Or, at least, some intuitive reasoning. (By the way, I notice in your first sentence that you don't mention definability, where I actually use this in my argument. Is this not necessary for the nonexistence of a strong master condition?)
Mar 12, 2023 at 9:55 comment added Monroe Eskew Yes, lifted through G.
Mar 12, 2023 at 9:13 comment added Connor W To clarify, at the end, when you say that $q$ is a strong master condition for $j$ and $\mathbb{Q}$, is this $j$ the lifted embedding?
Mar 12, 2023 at 9:01 history answered Monroe Eskew CC BY-SA 4.0