Tetration of 2 and Aleph_0
#1
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
#2
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.
#3
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
#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
#5
Darn it- my code didn't work. Can anybody show me the correct formatted code for some statements I just made?

Thanks,
Hassler
#6
(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:
[tex]\aleph_0[/tex]
\( \aleph_0 \)

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
#7
(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

#8
personally i reject ordinals , as you might have read elsewhere.

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

tommy1729
#9
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
#10
(09/07/2011, 03:37 AM)tommy1729 Wrote: 2^(aleph_aleph_0) = aleph_aleph_0

notice aleph_0 + 1 = aleph_0

and 2^^(aleph_aleph_0) = aleph_aleph_0
It is not true that [Image: png.image?\dpi%7B110%7D%202\uparrow\aleph_\o...eph_\omega].
It is also not true that [Image: png.image?\dpi%7B110%7D%202\uparrow\uparrow\...eph_\omega].
If the generalized continuum hypothesis is true, then [Image: png.image?\dpi%7B110%7D%202\uparrow\aleph_\o..._%7B\omega+1].
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\




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