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Jul 4, 2017 at 14:29 history edited Mikhail Katz CC BY-SA 3.0
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Jul 4, 2017 at 8:03 answer added Mikhail Katz timeline score: 1
Jun 20, 2016 at 13:30 comment added Mikhail Katz @JoelDavidHamkins, the definition is a formula in ZF. Otherwise one couldn't speak about a definable object at all. The magic trick is that in order to prove that the extension is actually proper you need more axiomatic material. In other words, models where there are no ultrafilters will simply give you the same field you started with.Notice that transfer happens to be true for the trivial extension :-)
Jun 20, 2016 at 13:19 comment added Joel David Hamkins Don't we expect models of ZF+ACC where the definition fails, simply by lack of ultrafilters?
Jun 20, 2016 at 12:07 history edited Mikhail Katz
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Jun 20, 2016 at 7:25 history edited Mikhail Katz CC BY-SA 3.0
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Jun 20, 2016 at 7:07 history asked Mikhail Katz CC BY-SA 3.0