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May 31, 2023 at 10:01 answer added C7X timeline score: 0
Jun 15, 2018 at 21:32 history edited godelian CC BY-SA 4.0
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Jun 15, 2018 at 17:21 review Close votes
Jun 16, 2018 at 0:34
Jun 15, 2018 at 16:58 comment added fedja "Also, being more reckless, we could generalize the above" When I hear something like that, I always ask "And what particular problem are we trying to solve with all this machinery here?". That is just my version of Occam's razor in its original form: "Entities should not be multiplied unnecessarily". On the other hand, I'm an old stupid guy who is already completely overwhelmed with elementary questions about continuous functions, convex sets, random walks on finite graphs, 5 variable inequalities, and other elementary school puzzles ;-)
Jun 15, 2018 at 16:53 history edited Qfwfq CC BY-SA 4.0
edited title
Dec 16, 2017 at 7:58 comment added Zetapology @MircoA.Mannucci What you are doing is precisely including $\alpha$-th order logic for transfinite $\alpha$ (or even just $\alpha>1$). In either case, what is achieved is a collection which has no bijection to $V$.
Jan 30, 2016 at 8:32 comment added Morteza Azad @AsafKaragila Any reference for where Cantor used Tav?
Jan 30, 2016 at 7:48 comment added Morteza Azad @NoamD.Elkies Regarding using Hebrew Letters in LaTeX, this post in Tex.SE could be of your interest.
Jul 2, 2012 at 6:14 comment added Noam D. Elkies Actually 27 Hebrew letters if you count final forms of KMNPTz. But only the first four are, um, accessible in LaTeX: you can't go past \daleth = $\daleth$.
Jul 2, 2012 at 6:05 answer added David Roberts timeline score: 4
Jul 1, 2012 at 16:52 history edited Mirco A. Mannucci CC BY-SA 3.0
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Jun 30, 2012 at 22:14 answer added Asaf Karagila timeline score: 6
Jun 30, 2012 at 20:07 answer added user24527 timeline score: 5
Jun 30, 2012 at 18:14 answer added Gerhard Paseman timeline score: 1
Jun 30, 2012 at 16:57 answer added Benedict Eastaugh timeline score: 10
Jun 30, 2012 at 11:45 comment added Asaf Karagila Actually Cantor used Tav (the last letter) as well.
Jun 30, 2012 at 11:27 history edited Mirco A. Mannucci CC BY-SA 3.0
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Jun 30, 2012 at 11:22 history edited Mirco A. Mannucci CC BY-SA 3.0
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Jun 30, 2012 at 11:16 comment added Mirco A. Mannucci message received. The hebrew alphabet has exactly 22 letters (autiot), so I would quickly run out of hyperinfinites, not to mention that, as Asaf and you have pointed out, Beth and Ghimel are already in use, for other purposes. I have a better idea (see my editing above). Thanks to you both for your good notational criticism! PS for Andreas: I will comment to your great answer later today: right now I need to recharge the solar batteries with a long walk in the woods, pondering on the new alephs
Jun 30, 2012 at 10:51 comment added Andreas Blass A continuation of Asaf's comment: The next Hebrew letter, gimel $\gimel$, is also already in use. But, as far as I know, he rest of the Hebrew alphabet is available.
Jun 30, 2012 at 7:03 comment added Asaf Karagila I should mention that $\beth$ numbers already exist in set theory. Using $\beth_0$ is a bad form of overloading.
Jun 30, 2012 at 0:36 answer added Joel David Hamkins timeline score: 28
Jun 30, 2012 at 0:26 answer added Noah Schweber timeline score: 6
Jun 30, 2012 at 0:25 answer added Andreas Blass timeline score: 22
Jun 30, 2012 at 0:18 comment added Henry Towsner Don't the existing notions of large cardinals already do this? Let $M_1$ is a model of ZFC+"there is an inaccessible", and let $M_0$ consist of those sets of size hereditarily smaller than the least inaccessible of $M_1$. This seems to be precisely the situation you describe. The "small" (sub-measurable) large cardinal notions are then the sorts of chains of increasing models of ZFC you describe.
Jun 29, 2012 at 23:33 history edited Mirco A. Mannucci CC BY-SA 3.0
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Jun 29, 2012 at 23:17 history asked Mirco A. Mannucci CC BY-SA 3.0