Timeline for The transcendence degree of $\mathbb R$ after adding a Cohen
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 6, 2014 at 12:55 | comment | added | Andrej Bauer | You should make sure to hit 100k during the Vienna summer logic meetings. We'll buy you a Sachertorte. | |
| May 6, 2014 at 10:50 | comment | added | Joel David Hamkins | @AndrejBauer I certainly hope so! But we're all far flung. On a separate note, I wonder if someday we might organize some kind of MO conference or perhaps an MO special session at one of the big AMS meetings. | |
| May 6, 2014 at 9:39 | comment | added | Andrej Bauer | Will there be a party when Joel hits 100k points? | |
| Apr 2, 2013 at 12:38 | comment | added | Asaf Karagila♦ | Err, yeah. My train of thoughts was disrupted by a phone and it seems that I continued my comment in an incompatible way to its beginning. I was trying to suggest that when reducing to the symmetric extension we exclude many real numbers from the universe. So perhaps the remainder of is small enough to be algebraically generated by the "canonical generics" (or some derivation of them), which would result in a transcendence degree which is either countable, or contains an infinite Dedekind-finite set (but not continuum itself). Your update seems to suggest otherwise. Which is good, thanks! | |
| Apr 2, 2013 at 11:49 | comment | added | Joel David Hamkins | Could you explain your comment? Without AC, does the concept of transcendence degree make sense? After all, I think it is possible that there is no transcendence base. If you mean a model something like $V(\mathbb{R}^{V[c]})$, then the argument of my update paragraph shows that there are sets of size continuum of mutually transcendental reals in the extension, which seems related. | |
| Apr 2, 2013 at 11:40 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 656 characters in body; added 67 characters in body
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| Apr 2, 2013 at 10:43 | vote | accept | Asaf Karagila♦ | ||
| Apr 2, 2013 at 10:43 | comment | added | Asaf Karagila♦ | That's great! Now I wonder what happens when we reduce ourselves to Cohen's first model's analogue (without requiring $V=L$ that is). I suspect that the transcendence degree there won't even be will either be $\omega$ or infinite Dedekind-finite cardinal $+\omega$. | |
| Apr 2, 2013 at 10:26 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |