Timeline for Can models of set theory contain extra ordinals?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2010 at 18:09 | vote | accept | Mike Shulman | ||
| Feb 23, 2010 at 18:09 | vote | accept | Mike Shulman | ||
| Feb 23, 2010 at 18:09 | |||||
| Feb 23, 2010 at 15:25 | comment | added | Joel David Hamkins | Mike, yes, that condition is exactly what it takes to run the argument I gave above. Your condition on ordinals is just a special case of it, unless z is an ordinal completely on top of the V ordinals. | |
| Feb 23, 2010 at 15:10 | comment | added | Joel David Hamkins | Yes, they are cofinal. | |
| Feb 23, 2010 at 15:08 | comment | added | Mike Shulman | Ah, looks like maybe you already answered this question in the comments to Francois -- I was going to ask, in your last example, is it still true that the ordinals of the form j(a) are cofinal in the ordinals of M? | |
| Feb 23, 2010 at 15:06 | comment | added | Mike Shulman | Thanks! I did have some expectation that this condition would allow names of ordinals whose "ordinal-ness" is "spread out over more than one ordinal" and thus not assertably equal to any particular one, but I think your last example is the one I was really looking for. I didn't state all of Blass & Scedrov's other requirements, but I'm guessing that that example also violates their condition defining the inclusion of V in M, namely $[[z \in \check{x}]] = \bigvee_{y\in x} [[z = \check{y}]]$ for all x in V. Is that right? | |
| Feb 23, 2010 at 14:22 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Added counterexample at end.
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| Feb 23, 2010 at 13:45 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |