Timeline for Large Cardinals Imply a Model of ZFC
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2013 at 0:22 | review | Suggested edits | |||
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| Jul 25, 2013 at 7:10 | review | Suggested edits | |||
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| Jul 23, 2013 at 10:12 | review | Suggested edits | |||
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| Jul 23, 2013 at 9:50 | review | Suggested edits | |||
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| Jul 23, 2013 at 2:42 | review | Suggested edits | |||
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| Jul 20, 2013 at 15:02 | review | Suggested edits | |||
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| Jul 19, 2013 at 15:40 | review | Suggested edits | |||
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| Jul 17, 2013 at 23:39 | review | Suggested edits | |||
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| Jul 9, 2013 at 22:11 | review | Suggested edits | |||
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| Jul 9, 2013 at 10:49 | review | Suggested edits | |||
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| Jul 9, 2013 at 8:19 | review | Suggested edits | |||
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| Jul 8, 2013 at 6:02 | review | Suggested edits | |||
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| Aug 15, 2011 at 12:29 | comment | added | Asaf Karagila♦ | Joel, Todd: Very well. I was not aware of this social convention, and the post begins with "I wanted to post a comment, but I don't have enough reputation.", now ftonti has enough reputation so the next time he could choose to write an answer or leave a comment. :-) | |
| Aug 15, 2011 at 12:08 | comment | added | Todd Trimble♦ | I agree with Joel. So, +1 to ftonti. | |
| Aug 15, 2011 at 11:58 | comment | added | Joel David Hamkins | Asaf, it is fine for ftonti to post a comment like this, when lacking sufficient rep to post a proper comment; my understanding is that our MO policy specifically allows this, and when the comment is worthwhile, as here, then this is often a good way to earn the necessary reputation. | |
| Aug 15, 2011 at 11:47 | comment | added | Andrés E. Caicedo | Yes, this has been pointed out a few times in answers to other questions here at MO: The existence of an $\omega$-model, i.e., one whose $\omega$ is isomorphic to the $\omega$ of the underlying universe, implies Con(ZFC), Con(ZFC+Con(ZFC)), Con(ZFC+Con(...)), etc. In turn, the existence of a transitive model implies the existence of an $\omega$-model, and an omega-model of "there is an $\omega$-model" etc. The existence of a transitive model is strictly weaker than the existence of a $\lambda$ such that $V_\lambda$ is a model, and this is strictly weaker than the existence of an inaccessible. | |
| Aug 15, 2011 at 11:42 | comment | added | Asaf Karagila♦ | Why not wait a bit more until you have the proper reputation to comment instead? You're not that far away... | |
| Aug 15, 2011 at 11:39 | history | answered | ftonti | CC BY-SA 3.0 |