Timeline for Is it possible to define higher cardinal arithmetics
Current License: CC BY-SA 3.0
8 events
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| Nov 25, 2014 at 4:21 | comment | added | Mohammad Golshani | For, the question, you might be right, but when I was asking it, I had essentially the same idea as your question. This is why I mentioned other operations before asking. Note that for example, $f_+(\kappa, \lambda)=|\kappa\times\{0\} \cup \lambda\times\{1\}|, f_\times(\kappa,\lambda)=|\kappa\times\lambda|,$ .... | |
| Nov 25, 2014 at 4:13 | comment | added | Noah Schweber | @MohammadGolshani, I think that tetration has a recursive character which exponentiation lacks. Obviously this is informal, but I don't find the fact that exponentiation can't be defined similarly to necessarily be a point against this. (Re: question 2, I don't think they're the same; one can have a natural operation corresponding to tetration - even one with a definition that lifts seamlessly to infinite cardinals - without having a concrete picture of the set being counted.) | |
| Nov 25, 2014 at 4:00 | comment | added | Mohammad Golshani | Also your last question is essentially my question 2. | |
| Nov 25, 2014 at 3:59 | comment | added | Mohammad Golshani | @NoahS In the above, you first define tetration for ordinals, and then use it to define tetration for cardinals, as a sup. To see if it natural, we might expect the same works for earlier cardinal operations, say exponentiation. But I don't see if we have$2^{\aleph_0}=\aleph_0^{\aleph_0}=sup\{\aleph_0^{\alpha}: |\alpha|=\aleph_0\},$ where in $sup$, ordinal exponentiation is considered. | |
| Nov 24, 2014 at 23:29 | comment | added | Noah Schweber | D'oh - of course you're right. Still, I find this specific example of discontinuity a little disconcerting. (Although right now I do think Approach 2 is the right way to define cardinal tetration.) | |
| Nov 24, 2014 at 23:28 | history | edited | Noah Schweber | CC BY-SA 3.0 |
added 30 characters in body
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| Nov 24, 2014 at 22:15 | comment | added | Andrés E. Caicedo | Cardinal exponentiation is already discontinuous, so why would this be considered a drawback? | |
| Nov 24, 2014 at 19:47 | history | answered | Noah Schweber | CC BY-SA 3.0 |