# Tetration Forum

Full Version: Tetration of 2 and Aleph_0
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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
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.
VERY controversial subject.

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

Thanks,
Hassler
(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
(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 $\text{slog}(\aleph_{\aleph_1})=\aleph_1$
- Shel
It 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

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

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

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