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Apr 23, 2013 at 12:58 comment added Joel David Hamkins It is also possible to do it with two countable models, as long as $N$ has a branch that is not in $M$.
Apr 23, 2013 at 12:01 comment added Cubikova This is very illustrative, thanks. Worth noticing that it is almost the same idea, since it also makes use tacitly of the same cardinality argument when we pick $M$ to be a countable substructure of $N$.
Apr 22, 2013 at 12:35 history edited Joel David Hamkins CC BY-SA 3.0
Fixed typo
Apr 22, 2013 at 12:22 history edited Joel David Hamkins CC BY-SA 3.0
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Apr 22, 2013 at 12:16 history answered Joel David Hamkins CC BY-SA 3.0