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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

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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

actually meant two times two

amount of times.

But as far as I know there isn't much research into tetrating

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.

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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

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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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Darn it- my code didn't work. Can anybody show me the correct formatted code for some statements I just made?

Thanks,

Hassler

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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

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?

which implies

And for any integer n where

, then

Perhaps

- Shel

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09/09/2011, 05:54 PM
(This post was last modified: 09/09/2011, 08:23 PM by sheldonison.)

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personally i reject ordinals , as you might have read elsewhere.

i feel inaccessible ordinals are far away from tetration btw...

tommy1729

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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