Aleph^aleph

**Phrak** · Jul4-09, 01:47 PM

A set of cardinality $LaTeX Code: \\aleph_0$ has elements that are sets of size $LaTeX Code: \\aleph_0$ , and so on.

Counting elements, I get

$LaTeX Code: \\aleph_0 + \\aleph_0^2 + ... + \\aleph_0^{\\aleph_0}\\ .$

Is this the same size as $LaTeX Code: \\aleph_0$ ?

**Hurkyl** · Jul4-09, 02:03 PM

Originally Posted by Phrak

A set of cardinality $LaTeX Code: \\aleph_0$ has elements that are sets of size $LaTeX Code: \\aleph_0$ , and so on.

I don't know what you mean by this

$LaTeX Code: \\aleph_0 + \\aleph_0^2 + ... + \\aleph_0^{\\aleph_0}\\ .$

What you're actually summing here isn't perfectly clear. But, at least, I can point out that $LaTeX Code: \\aleph_0^{\\aleph_0} > \\aleph_0$ .

**Phrak** · Jul4-09, 11:46 PM

I'm afraid I've blundered the original question.

A set A has cardinality of c, where c is the size of the set of real numbers. |A| = c.

B_r are the elements of A. The B_r are sets, themselves, of cardinality c.
C_r are the elements of each of the sets B_r. The C_r are sets, themselves, of cardinality c.
This continues, ad infinitum.

Counting elements the sum of the elements of every set, I get

x = c + c² + c³ + ...

Is x equal to c or larger?

**Hurkyl** · Jul5-09, 12:09 AM

Originally Posted by Phrak

x = c + c² + c³ + ...

Is x equal to c or larger?

Equal. Because:

$LaTeX Code: x = \\sum_{n \\in \\mathbb{Z}^+} c^n<BR> = \\sum_{n \\in \\mathbb{Z}^+} c<BR> = \\aleph_0 \\cdot c = c$

**Phrak** · Jul5-09, 12:38 AM

Originally Posted by Hurkyl

Equal. Because:

$LaTeX Code: x = \\sum_{n \\in \\mathbb{Z}^+} c^n<BR> = \\sum_{n \\in \\mathbb{Z}^+} c<BR> = \\aleph_0 \\cdot c = c$

Thanks Hykyl. Since time I asked, I found a raft of information in wiki's Cardinality section as well.

**Preno** · Jul5-09, 07:57 AM

Originally Posted by Phrak

I'm afraid I've blundered the original question.

A set A has cardinality of c, where c is the size of the set of real numbers. |A| = c.

B_r are the elements of A. The B_r are sets, themselves, of cardinality c.
C_r are the elements of each of the sets B_r. The C_r are sets, themselves, of cardinality c.
This continues, ad infinitum.

No, it doesn't, because of the axiom of regularity.

**Phrak** · Jul5-09, 11:20 AM

Originally Posted by Preno

No, it doesn't, because of the axiom of regularity.

What doesn't? I don't understand.

**Preno** · Jul5-09, 11:34 AM

The axiom of regularity says that you cannot have an infinite descending chain of elements, i.e. a sequence of sets

such that $LaTeX Code: x_{i+1} \\in x_i$ . So you cannot "continue ad infinitum".

**Dragonfall** · Jul5-09, 03:27 PM

If we're talking about ZFC.

**Phrak** · Jul5-09, 07:41 PM

Originally Posted by Preno

The axiom of regularity says that you cannot have an infinite descending chain of elements, i.e. a sequence of sets such that $LaTeX Code: x_{i+1} \\in x_i$ . So you cannot "continue ad infinitum".

Thanks for bringing this up. What is the reason this is contained in ZFC?

**Phrak** · Jul6-09, 12:04 AM

The idea is like taking a thing, dividing it into N pieces, dividing the pieces into N pieces, and so on, without end. Does ZFC set theory really preclude this?

**Preno** · Jul6-09, 07:34 AM

Yes. The motivation for the ZF axioms is the "iterative concept of set", wherein you build sets from the bottom up by iterating the power set and union, in stages, starting with the empty set. In ZF, it is equivalent to transfinite epsilon induction, i.e. it gives you a mechanism for proving stuff about all sets if you know that "all elements of x have P" implies "x has P" (notice the initial step takes care of itself automatically).

For an explanation of the iterative concept of set, see for example Parsons' What Is The Iterative Conception of Set? and Boolos' The iterative conception of set.

There are also "non-well-founded" set theories which violate the axiom of regularity/foundation. IIRC they are supposed have some use in computer science. Peter Aczel studied those, see his Non-well-founded Sets (which should be available online).

**Dragonfall** · Jul6-09, 01:38 PM

Pretty much the only use for the foundation axiom in ZFC is to prove that set membership is well founded, and iirc, if large cardinals exist, that V=L, Godel's constructible universe.

The main problem with non-well-founded sets is to define a notion of "equality". That and relative consistency results. To accomplish the former Aczel uses a notion borrowed from computer science, that of "bisimulation".

But even if you go with non-well-founded sets, there are no infinite descending membership chains longer than $LaTeX Code: \\omega$ .

WRT your first post, I think what you're asking is that if you have a set of sets, what is the size of its $LaTeX Code: \\in$ -closure?

**Phrak** · Jul12-09, 11:37 AM

I found this statement in wikipedia,

"However, no axiom system in first order logic [such as ZFC] is strong enough to fully (categorically) describe infinite structures such as the natural numbers or the real line. Categorical axiom systems for these structures can be obtained in stronger logics such as second-order logic,"

under first order logic.

Is it true? If so, I don't think ZFC applies to sets having size of |R|, or aleph_0 for that matter.

**Dragonfall** · Jul13-09, 10:34 AM

I think the keyword here is "fully (categorically)". I'm not sure what that means.

**Hurkyl** · Jul13-09, 05:07 PM

You've run into one of the subtleties that can be tricky about formal logic. The point is that you've overloaded the term "cardinality" -- you're using it to refer both to the notion of cardinality defined by whatever "ambient" mathematical system you're using, and you're using it to refer to the notion of cardinality as defined by the formalization of ZFC you're studying.

Confusing the two notions of cardinality can lead to all sorts of problems -- one of the most famous is Skolem's paradox.

I've referenced wikipedia as a starting point, but do not construe that as an endorsement of the article.