Timeline for Is it known whether every $\omega$-tree with an infinite antichain has an infinite chain in $\mathsf{ZF}$?
Current License: CC BY-SA 4.0
21 events
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| Nov 18, 2021 at 0:11 | comment | added | Cameron Buie | @Pace: Now that I think about it a little more, the example of usage provided by the dictionary was: "The statement is either true or false." Exclusivity is rather clear in that choice, so I gather that my understanding of the word "alternatives" (and of the dictionary-provided definition for the word "alternatives") was correct. Given that, I'd posit that your point amounts to the observation of a very common error. | |
| Nov 2, 2021 at 17:53 | comment | added | Cameron Buie | @Pace: That's a fair point of view. People use it inconsistently enough that there isn't really a clear common usage in conversational English, and even in mathematical English, there is some disagreement on the point, but most sources I've encountered use "either...or" to indicate exclusive disjunction (with this paper as the notable exception). Merriam Webster's definition of the conjunction "either" states the it is intended to precede a list of two or more alternatives, which I take to indicate exclusivity, but your mileage may vary. | |
| Nov 2, 2021 at 14:26 | comment | added | Pace Nielsen | @CameronBuie Wanted you to know that the word "either" is not incorrect, and doesn't need a "sic". The word "either" does not necessarily imply an "exclusive or", in either regular English or in mathematical English (or both!). It is often used just to signify that an "or" is coming. | |
| Nov 2, 2021 at 12:49 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
changed mobile Wikipedia link to the full site
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| Oct 28, 2021 at 17:34 | answer | added | Cameron Buie | timeline score: 6 | |
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Sep 27, 2017 at 14:22 | history | edited | YCor |
edited tags
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| Sep 27, 2017 at 13:12 | comment | added | Cameron Buie | @Martin: Yeah, that confused me, too. I just flipped a mental coin and went with one. | |
| Sep 27, 2017 at 13:11 | comment | added | Martin Sleziak | MathSciNet lists the author mentioned in the last paragraph as Keremedis - Keremedis, Kyriakos On infinite trees without infinite chains or antichains. Math. Japon. 51 (2000), no. 2, 175–178. ams.org/mathscinet-getitem?mr=1747289 I was about to edit the post, since I thought it was a typo - then I noticed that on his own website the same person uses name Keremidis. | |
| Sep 27, 2017 at 12:43 | history | edited | Martin Sleziak |
(infinite-combinatorics) seem like a suitable tag for me - feel free to revert my edit if you think it does not fit
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| Sep 27, 2017 at 12:40 | history | edited | Cameron Buie | CC BY-SA 3.0 |
Added definitions and clarified question.
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| Sep 27, 2017 at 12:20 | comment | added | Cameron Buie | @bof: Not at all. I'll do so now. | |
| Sep 27, 2017 at 5:39 | comment | added | bof | Would there be any harm in making this question accessible to the masses by including definitions of $\omega$-tree and antichain? | |
| Sep 26, 2017 at 23:51 | history | edited | Cameron Buie | CC BY-SA 3.0 |
updated, based on further research
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jan 15, 2014 at 17:32 | history | edited | Cameron Buie | CC BY-SA 3.0 |
deleted 8 characters in body
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| Sep 29, 2013 at 20:49 | comment | added | Noah Schweber | That's a nice proof. | |
| Sep 29, 2013 at 19:04 | comment | added | Cameron Buie | Finally, since $\mathcal A$ is well-orderable and each $|A|$ a finite cardinal, then $$\sum_{A\in\mathcal A}|A|=\max\left\{|\mathcal{A}|,\sup_{A\in\mathcal A}|A|\right\}\le\aleph_0.$$ | |
| Sep 29, 2013 at 19:02 | comment | added | Cameron Buie | @Noah: Let $\mathcal A$ be a countable set of finite sets. For each $A\in\mathcal A,$ we have that $A$ is well-orderable, so is in bijection with a unique ordinal--namely $|A|$. There are only finitely-many functions $A\to|A|,$ so the set $B_A$ of bijections $A\to|A|$ is finite and non-empty for each $A\in\mathcal A$. Then we can choose $g_A\in B_A$ for each $A\in\mathcal A$ since $\mathcal A$ is countable. We can readily show that $|\bigcup\mathcal A|\leq\sum_{A\in\mathcal A}|A|$ using these bijections. (cont'd) | |
| Sep 29, 2013 at 18:47 | comment | added | Noah Schweber | Probably a silly question, but: what is the proof that (3) implies (1)? | |
| Sep 29, 2013 at 18:26 | history | asked | Cameron Buie | CC BY-SA 3.0 |