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[MO] Is there a tetration for infinite cardinalities? (Question in MO)
(12/07/2014, 05:32 PM)tommy1729 Wrote: Long ago , I was challenged on sci.math to answer my own question.

There cannot be a bijection from exp^[1/2](w) to w nor 2^w.

The question itself is enough interesting. Assuming that the axiom of choiche holds over the Universe of set, we know that every set is well-orderable thus we can use the ordianl numbers (a sequence of transitive set) as a "point of reference", a unit of measure, to measure the size of every set...

So we are searching for a function over the cardinals (a special kind of ordinal numbers) such that ... is that right?
We should have that is the half iterate of the cardinal exponentiation.

The biggest problem here is that cardinal exponentiation isnt really defined... what i mean is that every set should have a cadinality that is an aleph number but actually we cant really say wich aleph number it is... I mean that has cardinality but we do not know what is the cardinality of the exponentiation of that cardinal number: has unknown cardinality... we do not know what is its size... that's the porblem behind CH...


So if there is not a cardinal number between and (in other words CH holds) how we can define a function over the cardinals number that is the half iterate?

from an half iterate we would like to have

or we don't?
If CH holds then there is not a cardinal strictly between omega and the continuum

So yes, if CH holds what you say is true

Quote:SO set theory is not suitable for functions since only the power function 2^ is a function that exists and makes a "difference".

( Polynomials do not make a "difference" , they do not change cardinality )

Considering that infinite set theory is not suitable for functions , it makes more sense to be skeptical about its use in other branches of math such as number theory , algebra , calculus and dynamical systems.

Also notice that even substraction is not well defined (for ordinals) since w-1 does not even exist.
I can't follow you because only if the functions over the (infinite)cardinals are counterintuitive (what di you expect from the infinity?) this doesn't mean that set theory is not suitable for functions at all.
Actually the polinomials, the exp function can be defined set-theoretically as a function over the real numbers...that can be modelled as a set (via dedekind cuts, or sequences of integers)...

Alot of topological concepts like continuity and stuff can be modelled set-teoretically. Alot of algebra too, like cosets, quotient structures, product structures, order theory... everything because we can define n-ary relations and n-ary functions over sets...
We can define rationals numbers Q as the cartesian product ZxZ up to an equivalence relation (quotient set)... so i think that is interesting set theory, and sometimes helps to makes thing clear because it give us a big workspace.

I don't know how many problem has ZFC, or in general the estensional set theories...probably alot, but they have positive sides as well. The fact that actually alot (98%?) of mathematical concept can be modelled by a sets that behave like the wanted object is impressive imho and outlines a little bit of effectiveness and charm in the set teoretical point of view...

Quote:That should shed some light on my skeptisism.
( it is JUST the tip of the iceberg , there is way more reason to be skeptical )

Since half-iterates (functions) do usually not exist in set theory , I do not see how one can continue to combine set theory and dynamics.



" Truth is what does not go away when you stop believing in it "

I don't know enough about dynamics so I don't want to continue, but we are talking about funnction over an exotic domain...the infinite cardinals... so it seems to me obvious that not everything that works for the function over the real numbers (a well known domain) should work on the extended domain of the higher infinities..

Thanks for the interesting topic btw.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
(12/01/2014, 11:55 PM)tommy1729 Wrote: 2^^w = the first inaccessible ordinal.

This is because 2^x = x is not accessible and 2^^w is the smallest such x.

w^^w = 2^(ln(w) 2^^w) = 2^^w = the first inaccessible ordinal.

2^^(2^^w) = the second inaccessible ordinal.

2^^(2^^w) = 2^^(w^^w) = w^^(w^^w).


w^^^w = the w th inaccessible ordinal.




the master
[Image: png.image?\dpi%7B110%7D%20\omega\uparrow\upa...\epsilon_0]. [Image: png.image?\dpi%7B110%7D%20\omega\uparrow\eps...\epsilon_0]. So [Image: png.image?\dpi%7B110%7D%20\omega\uparrow\upa...\epsilon_0].
Each time you add one to the thing you do omega to the power of it does not change, so [Image: png.image?\dpi%7B110%7D%20\omega\uparrow\upa...\epsilon_0].
[Image: png.image?\dpi%7B110%7D%20\omega\uparrow\upa...\epsilon_0].
Quote:Omega tetrated to the (omega plus one) being anything other than epsilon zero, such as epsilon one would not follow the recursive rule of tetration; that a tetated to the (b plus one) equals a to the power of a tetrated to the b.
That also causes [Image: png.image?\dpi%7B110%7D%20\omega\uparrow\uparrow\epsilon_0] and [Image: png.image?\dpi%7B110%7D%20\omega\uparrow\upa...\epsilon_0] to equal [Image: png.image?\dpi%7B110%7D%20\epsilon_0]. Even omega[omega]omega still equals equals epsilon zero.
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