A power set of infinite alephs

[Nate Lourwell]

Hey, all. I don't have much background yet in pure mathematics. In the meantime, I'm wondering if it would make any sense in set theory to make the set [math]kaph_{0}[/math] = {[math]n \in \mathbb{N}[/math] | [math]\aleph_{n}[/math]} and then make [math]\wp[/math]([math]kaph_{0}[/math]) = [math]kaph_{1}[/math], analogous to how [math]\wp[/math]([math]\aleph_{0}[/math]) = [math]\aleph_{1}[/math]. If that's possible, could it be continued without bound, next with [math]yodh_{0}[/math] = {[math]n \in \mathbb{N}[/math] | [math]kaph_{n}[/math]}?

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I learned that a correct way to describe [math]kaph_{0}[/math], if there is one, would be [math]kaph_{0}[/math] = {[math]\aleph_{i}[/math] | [math]i \in \mathbb{N}[/math]}. This set would fully capture the hierarchy of alephs, [math]yodh_{0}[/math] = {[math]kaph_{i}[/math] | [math]i \in \mathbb{N}[/math]} would fully capture the hierarchy of kaphs, and so on.

 

My inquiry is about whether that's sensibly possible or not and, if it is, about whether it's at least as mathematically meaningful as it intuitively appears.

[Nate Lourwell]

To continue coming up with new names for the sets would be difficult. The hierarchy of alephs is sometimes referred to as an ascending tower, so I selected [math]Tower[/math][0] = { [math]\aleph_{i}[/math] | [math]i \in \mathbb{N}[/math] } and, again, observed [math]\wp[/math]([math]Tower[/math][0]) = [math]Tower[/math][1] to grasp where it could lead. Without having to come up with a new name, there's [math]Tower[/math][1, 0] = { [math]Tower[/math][[math]i[/math]] | [math]i \in \mathbb{N}[/math] }, indicating a base-[math]\mathbb{N}[/math] positional system.

 

In general, [math]Tower[/math][[math]1 _{n+1}[/math], [math]0_{n}[/math], . . . , [math]0_{2}[/math], [math]0_{1}[/math]] = { [math]Tower[/math][[math]i_{n}[/math], . . . , [math]i_{2}[/math], [math]i_{1}[/math]] | ([math]i_{n}[/math], . . . , [math]i_{2}[/math], [math]i_{1}[/math]) [math]\in base[/math] [math]\mathbb{N}[/math] }.

Edited October 27, 2009 by Nate Lourwell

[Nate Lourwell]

It's stated on page 261 of Set Theory and Its Philosophy (Potter 2004), "The alephs do not form a set," with a proof. I'll have to try to understand it later, but this makes my set constructions an unlikelihood.

 

Although disappointing, I'm glad for a firm answer to my inquiry this early on.

[Abstract_Logic]

Hello Nate.

 

I haven't much experience with Set Theory either, but it is a very intriguing field of study.

 

From what I understand, the aleph numbers are used to represent the cardinality of sets of numbers, i.e. integers, natural numbers, etc. So it would make sense that the alephs cannot form sets themselves because, while they are objects of study within set theory, they are not subjected to the same operations as sets. Your proposal is more along the lines of something that would be called 'Meta-set theory'. And of course, this brings up the issues of the Continuum hypothesis and Russell's paradox. Let's say your proposal is plausible within mathematics, and you have a set of alephs, i.e. a power set of set cardinalities. Then, according to the methods of set theory, you would need a cardinality set for the power aleph set, and also a power-power aleph set for those sets, and so on. This is where things border on absurdity.

 

So your proposal, although very insightful, is unfortunately not plausible within Set theory.

 

But please, don't take my response as an end-to-it-all. I have very little background in set theory, so I think you should hold on to your notions and continue to ask around about them.

 

Good luck to you!

[Nate Lourwell]

You seem that you could independently come up with that proof yourself on page 261 in Set Theory and Its Philosophy (Theorem. The alephs do not form a set). It's about cardinality.

 

Best hopes for you as well!