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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 \( 2^{\aleph_0} \) 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.
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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
\( \aleph_0 \)
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
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09/09/2011, 05:54 PM
(This post was last modified: 09/09/2011, 08:23 PM by sheldonison.)
(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
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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
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06/19/2022, 09:55 AM
(This post was last modified: 08/15/2022, 06:01 AM by Catullus.)
(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
.
It is also not true that
.
If the
generalized continuum hypothesis is true, then
.
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\