Talk: Cardinal

Cardinal Inequalities

Interesting note, from here it can be shown that \(2^{\aleph_{\aleph_{0}}} < \aleph_{\aleph_{4}}\). —Preceding unsigned comment added by Kaptain Kerbab (talk • contribs)

What you said is actually wrong. Inequality \(2^{\aleph_\omega} < \aleph_{\omega_4}\) can be shown to be true under the assumption that \(\aleph_\omega\) is strong limit, i.e. that \(2^{\aleph_n}<\aleph_\omega\) for \(n<\omega\). This is one of the most famous results of PCF theory. LittlePeng9 (talk) 14:01, May 5, 2015 (UTC)

Bijections

How do you create a bijection from $\omega$ to $\omega^{\omega}$? Or, assuming CH, a bijection from $\omega$ to $\omega^{CK}_{1}$? King2218 (talk) 11:43, July 20, 2015 (UTC)

Without writing down an explicit bijection, we can easily show that one exists using Schröder–Bernstein theorem. First of all, there is obvious injection from \(\omega\) to both \(\omega^\omega\) and \(\omega^\text{CK}_1\). Now ordinals below \(\omega^\omega\) are of the form \(\omega^na_n+...+\omega a_1+a_0\), so there is injection from \(\omega^\omega\) to set of finite strings \((a_n,...,a_0)\) of natural numbers, and from these there is an injection to \(\omega\) given by \(2^{a_0}+2^{a_0+a_1}+...+2^{a_0+...+a_n}\).

For injection from CK ordinal to \(\omega\), every ordinal \(\alpha\) below CK ordinal is coded by some Turing machine, which under fixed encoding of TMs has number \(n_\alpha\). The injection is given by \(\alpha\mapsto n_\alpha\).

This proof at no point uses CH, and indeed CH has nothing to do with ordinals up to CK ordinal. LittlePeng9 (talk) 12:24, July 20, 2015 (UTC)

woops sorry for a moment there I thought that $\aleph_1$ could be less than or equal to $\omega^{CK}_{1}$ (and apparently it's not) King2218 (talk) 13:15, July 20, 2015 (UTC)

Exponentiation on aleph-zero

If \(2^{\aleph_0}\) exists and is unique, what about \(3^{\aleph_0}\)? Or \(5^{\aleph_0}\), ... \(13^{\aleph_0}\), \(212,938,732,199^{\aleph_0}\), \(\text{Graham's number}^{\aleph_0}\), \(\text{TREE(3)}^{\aleph_0}\), \(\text{Loader's number}^{\aleph_0}\), \(\text{Little Bigeddon}^{\aleph_0}\), etc.

All of these must be possible, mustn't they? And they must be different?

Or, while we're at it, what if the base is a non-integer, or better yet, \(\aleph_0\) itself?

Does there turn into infinitely many new exponential sets?

ArtismScrub (talk) 14:50, December 31, 2017 (UTC)

One can show all those are equal - for any finite cardinal \(n\), \(n^{\aleph_0}=\aleph_0^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}\). To get a larger cardinal, we can take for example \(2^{2^{\aleph_0}}\). LittlePeng9 (talk) 15:12, December 31, 2017 (UTC)

Cofinality section

The section on cofinality here uses cardinal addition, which while correct, isn't identical to the explanation used on the article Cofinality, nor is it identical to Jech's explanation, which would be a useful source. I propose to change it to the supremum definition C7X (talk) 20:12, 12 May 2021 (UTC)

I agree with it. Using addition is not so helpful for the reader to grasp the cofinality, and is less standard in the explanation.

p-adic 00:48, 13 May 2021 (UTC)

OK, I will change it C7X (talk) 01:09, 13 May 2021 (UTC)

What should I do with the \(\aleph_1\) section that I commented out? I can't think of a better way to word it in terms of limits, since it uses \(\aleph_0+\aleph_0+\aleph_0+\cdots\) C7X (talk) 01:54, 13 May 2021 (UTC)

Addition is not bad. The reason why it is good to replace the addition in the cofinality section by the limit is because the limit is clearer in the context. If addition is clearer in another context, then it is good to use it.

p-adic 14:52, 13 May 2021 (UTC)