Timeline for Does van der Waerden's Theorem hold for $\omega_1$?
Current License: CC BY-SA 3.0
18 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 22, 2015 at 20:10 | answer | added | Will Brian | timeline score: 3 | |
| May 22, 2015 at 19:51 | history | edited | Will Brian | CC BY-SA 3.0 |
I've stripped away all my previous edits and moved them into an answer.
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| May 22, 2015 at 18:19 | comment | added | Joel David Hamkins | Will, regarding your recent edits, it seems to me that the whole discussion would be easier to follow if in the question you stuck to the original original question only, and made all your other updates as a separate answer below, giving a clear summary as an answer all the further information that you've learned about it. | |
| May 22, 2015 at 17:26 | history | edited | Will Brian | CC BY-SA 3.0 |
I've added a summary of what we now know about this question.
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| May 19, 2015 at 15:10 | history | edited | Will Brian | CC BY-SA 3.0 |
added 11 characters in body
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| May 19, 2015 at 15:03 | history | edited | Will Brian | CC BY-SA 3.0 |
added a theorem
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| May 18, 2015 at 22:01 | comment | added | Will Brian | Thanks -- it was a very nice answer! I've been thinking about the natural sum and product for the last half hour, and I'm convinced that van der Waerden's Theorem holds in that case. I'm writing this now in a comment because I won't have time to write out a proof until tomorrow. | |
| May 18, 2015 at 21:57 | comment | added | Joel David Hamkins | Ah, you are right about $\alpha+F\cdot\beta$ making it trivial (even when $F$ is countably infinite). I want to think more about the natural sum and product case... By the way, it was a very nice problem! | |
| May 18, 2015 at 21:32 | comment | added | Will Brian | @Joel: If we use $\alpha + F \cdot \beta$ then choosing the right $\beta$ will make $F \cdot \beta$ a single point, which makes the conclusion trivial. Using the natural sum and product could be very interesting, though (I hadn't thought of that). I think (a modification of) the ultrafilter proof of van der Waerden's Theorem might go through in that case, but I'll need to double-check the details before I can say for sure. If it works out I'll let you know. | |
| May 18, 2015 at 21:00 | comment | added | Joel David Hamkins | It may be interesting to note that my argument really used that you were using the usual ordinal arithmetic (and also that you consider $\alpha+\beta\cdot F$ rather than $\alpha+F\cdot\beta$). If you have used the natural sum and product, instead of the usual sum and product, then it would break my counterexample. | |
| May 18, 2015 at 19:11 | history | edited | GH from MO |
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| May 18, 2015 at 18:48 | comment | added | Will Brian | @Ben, concerning your question about $\omega^\omega$, it looks like Joel's answer below applies equally well to $\omega^\omega$, or for that matter any indecomposable ordinal. So $\omega$ is the unique ordinal satisfying van der Waerden's Theorem. | |
| May 18, 2015 at 18:42 | vote | accept | Will Brian | ||
| May 18, 2015 at 18:27 | answer | added | Joel David Hamkins | timeline score: 19 | |
| May 18, 2015 at 16:04 | comment | added | Will Brian | I don't, but I think it's another good question. It's fairly easy to show that if it holds for some ordinal $\alpha$ then $\alpha$ must be indecomposable (a power of $\omega$). | |
| May 18, 2015 at 15:58 | comment | added | Ben Barber | Do you know if this is true in, say, $\omega^\omega$? | |
| May 18, 2015 at 15:20 | history | asked | Will Brian | CC BY-SA 3.0 |