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Oct 9, 2018 at 12:02 vote accept M Carl
Oct 9, 2018 at 10:21 answer added Monroe Eskew timeline score: 5
Oct 9, 2018 at 9:52 comment added M Carl You are right, the question was missing the assumption that $\alpha$ is countable in $L[a]$. Sorry for that.
Oct 9, 2018 at 9:50 history edited M Carl CC BY-SA 4.0
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Oct 8, 2018 at 19:47 comment added Yair Hayut I think that I don't understand the question, or that some assumptions are missing: if we take $\alpha = \beta$ to be a regular uncountable cardinal in $L$ then every maximal antichain in $P_\alpha$ in $L$ belongs also to $L_{\beta}[a]$. Thus, $L_\beta[a]$-generic is the same as $L$ generic. If $a$ cannot be added by the Cohen forcing then clearly $a\notin L[x]$ for any $L_\beta[a]$-generic filter $x$.
Oct 8, 2018 at 17:03 history asked M Carl CC BY-SA 4.0