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 Cardinality of Infinite tetration Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 06/18/2008, 08:12 AM (This post was last modified: 06/18/2008, 08:48 PM by Ivars.) andydude Wrote:Ivars Wrote:c=2^Aleph_0 This has not been proven. In fact this is the continuum hypothesis. Andrew Robbins The question is not so much about the continuum Hypothesis which may be true, may be not, but generally about cardinality of sets produced via operations, including cardinality of Number. E.g. If we have integer number N, its cardinality is assumed |N |which is just the number of elements in set {1,2,...,N}; However, when this N is produced as a result of multiplication of 2 sets representing n1 and n2 than the resulting set is multiset, so its cardinality is greater than |N|. E.g. If we have set {0,1} and {0,1} each of them has cardinality |2|.But product: {0,1}*{0,1} = {0*0, 0*1, 1*0,1*1} = {0,0,0,1} has cardinality |4| with 0 repeated 3 times. So my question arose from looking at exponentiation: If we have a number N2 such that N2= n1^n1 than its cardinality is also |n1^n1| as a result of such operation. It corresponds to choices of n1 from n1, or, combinatorically, if we have n1 differently labeled botlles of beer, how many ways (permutations) we can arrange them in n1 boxes with repetition and when order matters. If we now calculate n1^(n1^n1) the number of boxes goes up to n1^n1, number of permutations n1^(n1^n1), which is also the cardinality of Number N3 = n13 when obtained via tetration? Unfortunately, numbers in tetration are big, but Integers that can in principle have such cardinality and are possible to partition exponentially in 2 integers are not so many (if we exclude 1 as meaningless in exponentiation): 4=2^2, 8=2^3, 9= 3^2, 16=2^4=2^2^2, 25=5^2, 27=3^3, 32= 2^5, 36=6^2, 49=7^2, 64=2^6=4^3, 81=3^4=3^2^2 ,100=10^2, 125=5^3, 128=2^7, 216=6^3, 243=3^5, 256=2^8=2^2^3, 343=7^3, 512=2^9=2^3^2, 625=5^4, 729=3^6, 1024=2^10, 1296=6^4, 2048=2^11, 2187=3^7, 2401=7^4, 3125=5^5, .. Of these, only few can be partitioned in 3 integers with repetition: 16=2^2^2, 81=3^2^2, 256=2^2^3, 512=2^3^2, 625=5^2^2, 1296=6^2^2, 2401=7^2^2, Without repetition, we get 2 ways to partition some numbers infinitely : exponential factorials n^..4^3^2 and 3_factorials 2^3^....n. May be I have missed some. So the question I put forth in initial post is about cardinality of Numbers resulting from x2, xn, xoo. if x is integer, it is relatively simple to have a multiset as a result of exponentiation which includes all permutations . This can be used also in cases n^(n-1)^(n-2) leading to parts of exponential factorial, and n^(n+1)^(n+2) leading to parts of 3_factorial. If x is fractional, it requires interpretation of irrational cardinality, as e.g. (1/2)^(1/2)= sqrt(2) if x is irrational or transcendental or imaginary, such cardinality gets very strange. E.g. such cardinality of i^i is e^(-pi/2). The only place I have found any ideas in that direction so far is Euler characteristic of categories (and polyhedral sets) which can be imaginary, negative and fractional(see Leistner, Baez, Propp): Propp. Euler measure as generalized cardinality Ivars « Next Oldest | Next Newest »

 Messages In This Thread Cardinality of Infinite tetration - by Ivars - 06/17/2008, 01:02 PM RE: Cardinality of Infinite tetration - by Ivars - 06/17/2008, 07:59 PM RE: Cardinality of Infinite tetration - by andydude - 06/17/2008, 08:52 PM RE: Cardinality of Infinite tetration - by Ivars - 06/18/2008, 08:12 AM RE: Cardinality of Infinite tetration - by bo198214 - 06/20/2008, 02:16 PM RE: Cardinality of Infinite tetration - by Ivars - 06/20/2008, 04:28 PM RE: Cardinality of Infinite tetration - by bo198214 - 06/20/2008, 09:18 PM RE: Cardinality of Infinite tetration - by Ivars - 06/21/2008, 07:27 PM RE: Cardinality of Infinite tetration - by Ivars - 07/13/2008, 08:07 PM

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