Tetration of 2 and Aleph_0 jht9663 Junior Fellow Posts: 3 Threads: 1 Joined: Aug 2011 09/06/2011, 03:47 PM Assuming ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice), the continuum hypothesis proposes that 2^Aleph_0 = Aleph_1. Does anyone have any insight into the tetration of 2 and Aleph_0 ? I have no idea as to where to start on this problem. But I feel that it is important because it could lead to the recognition of new types of infinities. Also, please excuse my lack of formatting skills. I would greatly appreciate any help in producing formatted code. Thanks, Hassler Thurston JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 09/06/2011, 09:28 PM (This post was last modified: 09/06/2011, 09:34 PM by JmsNxn.) I always thought that was just convenience of notation for some other set operation; I didn't know $2^{\aleph_0}$ actually meant two times two $\aleph_0$ amount of times. But as far as I know there isn't much research into tetrating $\aleph_0$ And to produce code you'll need to learn Latex, it's a rather simple html-like code that most math forums have to format formulae. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/07/2011, 03:37 AM VERY controversial subject. many flamewars going on about this. my opinion is this 2^^aleph_0 = aleph_aleph_0 and further 2^(aleph_aleph_0) = aleph_aleph_0 notice aleph_0 + 1 = aleph_0 and 2^^(aleph_aleph_0) = aleph_aleph_0 notice 2 * aleph_0 = aleph_0 aleph_aleph_1 or higher does not exist. notice that defining what aleph_aleph_1 is the diagonal argument / powerset of is not possible ... ( which is imho required to assume existance of aleph_aleph_1 ) regards tommy1729 jht9663 Junior Fellow Posts: 3 Threads: 1 Joined: Aug 2011 09/07/2011, 03:33 PM So essentially [$\aleph_{\aleph_{0}}+1=\aleph_{\aleph_{0}}$], [$2*\aleph_{\aleph_{0}}=\aleph_{\aleph_{0}}$], [$2^\aleph_{\aleph_{0}}=\aleph_{\aleph_{0}}$], and 2^^[$\aleph_{\aleph_{0}}=\aleph_{\aleph_{0}}$]. However I do not agree that [$\aleph_{\aleph_{1}}] does not exist. My heuristic reasoning is: 1 (the first integer past the addition identity) + 0 = 1 (the first integer past 0) (assuming the Continuum Hypothesis) 2 (the first integer past the exponentiation identity) ^ [$\aleph_{0}$] = [$\aleph_{1}$] (1 being the first integer past 0) if these are true then 2 (the first integer past the pentation identity) ^^^ [$\aleph_{\aleph_{0}}$] = [$\aleph_{\aleph_{1}}$] (1 being the first integer past 0) and you could extend the pattern. Of course I have no other reasons to believe that the third statement is true, as one would have to prove that there does not exist a bijection from [$\aleph_{\aleph_{0}}$] to 2^^^[$\aleph_{aleph_{0}}\$]. Also, where would be a place I could go to on the internet to find more discussion on this topic? Thanks, Hassler Thurston jht9663 Junior Fellow Posts: 3 Threads: 1 Joined: Aug 2011 09/07/2011, 03:34 PM Darn it- my code didn't work. Can anybody show me the correct formatted code for some statements I just made? Thanks, Hassler sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 09/07/2011, 08:47 PM (This post was last modified: 09/07/2011, 09:05 PM by sheldonison.) (09/07/2011, 03:34 PM)jht9663 Wrote: Darn it- my code didn't work. Can anybody show me the correct formatted code for some statements I just made? Thanks, Hassler Try putting tex codes around your math statements Code:$$\aleph_0$$$\aleph_0$ I'm no expert on set theory, but on a humorous note (not mathematically sound), assuming the generalized continuum hypothesis, then what happens if we take the slog of an aleph number? $\aleph_1=2^{\aleph_0}$ which implies $\text{slog}_2(\aleph_1) = \text{slog}_2(\aleph_0)+1=\aleph_0$ And for any integer n where $\aleph_{n+1}=2^{\aleph_n}$, then $\text{slog}(\aleph_n)=\aleph_0$ Perhaps $\text{slog}(\aleph_{\aleph_1})=\aleph_1$ - Shel sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 09/09/2011, 05:54 PM (This post was last modified: 09/09/2011, 08:23 PM by sheldonison.) (09/07/2011, 08:47 PM)sheldonison Wrote: I'm no expert on set theory, but on a humorous note (not mathematically sound), assuming the generalized continuum hypothesis, then what happens if we take the slog of an aleph number? $\aleph_1=2^{\aleph_0}$ which implies $\text{slog}_2(\aleph_1) = \text{slog}_2(\aleph_0)+1=\aleph_0$ And for any integer n where $\aleph_{n+1}=2^{\aleph_n}$, then $\text{slog}(\aleph_n)=\aleph_0$ Perhaps $\text{slog}(\aleph_{\aleph_1})=\aleph_1$ - ShelIt turns out aleph and beth numbers should be indexed by ordinal numbers. The ordinal number equivalent to $\aleph_0=\omega$ and the ordinal number equivalent to $\aleph_1=\omega_1$ But I have no idea whether slog or sexp have any meaning for $\aleph$ numbers. The other possibility would be to see if sexp/slog would be more applicable to ordinal numbers. But the exponentiation rules for ordinal arithmetic say that $2^\omega=\omega$ I'm unsure of what $\text{sexp}(\omega)$ would be; the result might just be $\omega$. http://en.wikipedia.org/wiki/Ordinal_arithmetic http://en.wikipedia.org/wiki/Aleph_number tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 09/10/2011, 12:22 PM personally i reject ordinals , as you might have read elsewhere. i feel inaccessible ordinals are far away from tetration btw... tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 11/13/2011, 12:09 PM (This post was last modified: 11/13/2011, 12:54 PM by tommy1729.) the large cardinals and large ordinals are very axiomatic in nature. so without proofs of bijections or the lack of bijections it is pretty hard to talk about that. ( although i do like the comments here ) in my not so humble opinion its also a matter of taste because of the above and because of the possible use of ZF©. ( which has not been proven consistant ! ) i already commented my personal large cardinal axioms ( kinda ) , but i feel it is more intresting to consider small cardinalities. to be specific : what is the cardinality of f(n) where n lies between n and 2^n ? since cardinalities are not influenced by powers card ( Q ) = card ( Q ^ finite ) we can write our question as for n <<< f(n) <<< 2^n card(f(n)) = ? the reason i dont want to get close to n or 2^n is the question : is there a cardinality between n and 2^n ? in other words : the continuum hypothesis. in stardard math and standard combinatorics , we usually do not work with functions f : n <<< f(n) <<< 2^n. but on the tetration forum they occur very often. card(floor(sexp(slog(n)+1/(24+ln(ln(n)))))) = ? card(floor(n + n^4/4! + n^9/9! + n^16/16! + ...)) = ? regards tommy1729 « Next Oldest | Next Newest »