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Mar 20, 2018 at 17:53 vote accept Jérôme Poineau
Mar 20, 2018 at 13:34 comment added David Lampert The same basic argument works for arbitrary $k$: the set $\{f_i\}$ has the same cardinality as $k$ and same cardinality as $X(k)$, well-order $\{f_i\}$ by this cardinal, any proper initial segment of $\{f_i\}$ has smaller cardinality and so $P$ and $Q$ can be chosen for the next $f_i$.
Mar 20, 2018 at 12:30 comment added Jérôme Poineau Very nice, thanks! I am still interested in the general case though. I will edit the question.
Mar 19, 2018 at 23:26 review First posts
Mar 19, 2018 at 23:29
Mar 19, 2018 at 23:23 history answered Dragon CC BY-SA 3.0