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Tetration of 2 and Aleph_0
#4
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
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Messages In This Thread
Tetration of 2 and Aleph_0 - by jht9663 - 09/06/2011, 03:47 PM
RE: Tetration of 2 and Aleph_0 - by JmsNxn - 09/06/2011, 09:28 PM
RE: Tetration of 2 and Aleph_0 - by tommy1729 - 09/07/2011, 03:37 AM
RE: Tetration of 2 and Aleph_0 - by jht9663 - 09/07/2011, 03:33 PM
RE: Tetration of 2 and Aleph_0 - by sheldonison - 09/07/2011, 08:47 PM
RE: Tetration of 2 and Aleph_0 - by sheldonison - 09/09/2011, 05:54 PM
RE: Tetration of 2 and Aleph_0 - by jht9663 - 09/07/2011, 03:34 PM
RE: Tetration of 2 and Aleph_0 - by tommy1729 - 09/10/2011, 12:22 PM
RE: Tetration of 2 and Aleph_0 - by tommy1729 - 11/13/2011, 12:09 PM



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