(12/04/2014, 01:20 PM)tommy1729 Wrote: Doner and Tarski can not define exp^[1/2](w) and its cardinality , right ?The problem is not if they are able.. probably they weren't but as far as I know the didn't even tried, at least not in their work about the hyperoperations over transfinite ordinals.
The reason is clear, it was about transfinite ordinals not cardinals or fractional iteration... so your problem about the half iterate of the ordinal exponentiation has nothing to do with their work.
When you ask something like that, it should be clear, we have to be precise and define our words.
-What exp means?
-What half iterate means?
-what \( \omega \) means?
The answers to this questions involve a amount of non-trivial choiches like wich framework theory we should use, how to formalize the concept of infinity, potential or actual infinity? Cardinals or ordinals? How we define them? Axiom of choiche or not?
This mess of possible choices and interpretation makes your question blurry and ill defined even if I admit that most of the interpretations we can give to the question are really interesting.
Quote:Do Doner and Tarski give a different answer to 2^^w ?Sure, lets define first the ordinal tetration for limit ordinals
\( {}^\lambda\alpha=sup\{{}^\delta \alpha:\delta<\lambda\}=sup\{0;\alpha;{}^2\alpha;{}^3\alpha;...\} \)
So \( {}^\omega 2=sup \{{}^n 2:n<\omega\}=sup\{0,2;4;16;...\}=\omega \)
About the first fixed point of the function \( \alpha \mapsto \omega ^{\alpha} \)
or in other words the solution of \( \omega ^x=x \) we have that \( {}^\omega \omega \) is the first fixed point and is called epsilon zero
\( \epsilon_0={}^\omega\omega=sup\{{}^n \omega:n<\omega\}=sup\{0,\omega,\omega^{\omega},{}^3\omega,{}^4\omega, ...\} \)
The next fixed points can be defined recursively as follows for all the ordinals
\( \epsilon_{\delta+1}={}^\omega\epsilon_{\delta}=sup\{{}^n \epsilon_{\delta}:n<\omega\}=
sup\{ 0, \epsilon_{\delta}, \epsilon_{\delta}^{\epsilon_{\delta}}, {}^3\epsilon_{\delta}, {}^4 \epsilon_{\delta},...\} \)
for limit ordinals we have
\( \epsilon_{\lambda}=sup\{\epsilon_{\alpha}:\alpha<\lambda\} \)
Is possible to see that for the recursive definition we do not iterate the ordinal superexponentiation \( x \mapsto {}^x \alpha \) but we use the iteration of the map \( x \mapsto {}^\alpha x \) so the ordinal pentation is not involved at all with the fixed points.
Quote:I still challenge to give exp^[1/2](w) even under the assumption that ZFC is good.In my opinion the theory of surreal numbers can help us to extend the iteration of ordinal operations beyond the set theoretic ordinals (that are discrete, integer-like).
But is hard to think about the iteration of the Cardinal operations... really hard.
If I remember good the cardinals number are not even linear orderable without the axiom of choiche and the problem is that for infinite cardinals the cardinal exponentiation is linked more with the combinatorial constructions (like set of maps, powersets) rather than the recursive definitions.
So the key could be the combinatoral meaning of tetration (like the author of the questions seems to suggest).
Thinking about something simple I noteced that the beth numbers (I have to use the mathfrak symbol) are defined using iterated cardinal exponentiation.
\( \mathfrak B _0:=\aleph_0=\omega \)
\( \mathfrak B _{\alpha+1}:=2^{\aleph_0} \)
\( \mathfrak B _{\lambda}:=sup\{\mathfrak B _{\alpha}:\alpha<\lambda\} \)
the exponentiation is not the ordinal one, note that the cardinal exponentiation push us beyond the countable while the ordinal one (with the epsilon fixed point) can't go beyon the first uncontable \( \omega_1=\aleph_1 \)
Anyways we can define a kind of tetration that is defined for the cardinal base \( \aleph_0 \) and for transfinite ordinal superexponents
\( {}^1 \aleph_0=\mathfrak B _0=\aleph_0 \)
\( {}^2 \aleph_0=\aleph_0^{\aleph_0}=2^{\aleph_0}=\mathfrak B _1 \)
\( {}^{\alpha+1} \aleph_0=\aleph_0^{\mathfrak B _{\alpha}}=2^{\mathfrak B _{\alpha}}=\mathfrak B _{\alpha} \)
But about beth omega... that should be \( {}^\omega \aleph_0=\mathfrak B _\omega \) I'm not really sure it is also inaccessible... wikipedia says that it is the smallest strong limit cardinal but ZFC doesn't prove it (so I guess a fixed point of the cardinal exponentation) but there is the regularity requirement which I don't fully understand (it has to do with cofinalty but beth omega is the union of countable many smaller ordinal so i guess it is not inaccessible).
PS: I made i big mistake because strong limit doesn't mean that is a cardinal exp. fixed point but that for every smaller cardinal \( \kappa<{}^\omega \aleph_0 \) then \( 2^\kappa<{}^\omega \aleph_0 \). The fixed point statement is not compatible with the Cantor theorem.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
