Timeline for What choice principles does "every set is in bijection with a transitive set" imply?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 2 at 15:21 | comment | added | Joel David Hamkins | To my way of thinking, TC is similar in spirit to RR---it is a natural strengthening of RR. | |
| Oct 2 at 14:23 | comment | added | Joel David Hamkins | It lies exactly at RR. | |
| Oct 2 at 13:52 | comment | added | Asaf Karagila♦ | Joel, sure, it's a nice lower bound. But as you say, you don't really know where that lower bound actually lies in the hierarchy of choiceless principles... That TC is strictly above ZF was already noted in the question! | |
| Oct 2 at 13:21 | comment | added | Joel David Hamkins | I'm not sure about that. | |
| Oct 2 at 13:04 | comment | added | Will Brian | Nice answer -- +1. Do you know whether your lower bound improves on the one in the question? In other words, do you know whether it is consistent that RR fails, but still every set surjects onto $\omega$? | |
| Oct 2 at 12:52 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Oct 2 at 12:33 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Oct 2 at 12:31 | comment | added | Joel David Hamkins | It provides a lower bound, as requested. | |
| Oct 2 at 12:28 | comment | added | Asaf Karagila♦ | Sure, but AC implies RR as well... | |
| Oct 2 at 12:19 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Oct 2 at 12:14 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |