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I am starting to read the Kunen's book set theory. I could not understand the why we need cofinality. Why is it important?

Also, I'm trying to solve some exercises (on page 146) about absoluteness for $R(\kappa)$, where $\kappa$ is strongly inaccessible.

The exercise in question is the following (Exercise 2):

(AC) Let $\kappa$ be strongly inaccessible. Check that the following are absolute for $R(\kappa)$:

(a) $\mathcal{P}(x)$.

(b) $\omega_\alpha$.

(c) $\gimel_\alpha$.

(d) $R(\alpha)$.

(e) $\mathrm{cf}(\alpha)$.

(f) $\alpha$ is strongly inaccessible.

I think that $\mathcal{P}(x)$, $\omega_\alpha$ and the sequence $(\omega_\alpha)$ (which is defined by transfinite recursion on $\alpha$) are absolute for $R(\kappa)$ because all of them are transitive sets. But in (e) and in (f) I could not prove anything. Could you give me a hint?

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4  
If you only start with set theory Kunen is not the right choice. –  Asaf Karagila May 12 '11 at 13:06
    
What are (e) and (f)? –  Apostolos May 12 '11 at 14:00
1  
This is elementary set theory? –  Phira May 12 '11 at 14:44
    
Is it possible to give a title which is coherent with the question? –  Asaf Karagila May 12 '11 at 17:46
    
Is it really $\gimel$ rather than $\beth$ in (c)? –  Yuval Filmus May 12 '11 at 22:18

2 Answers 2

up vote 15 down vote accepted

If you are just starting with cofinality, then it will be illumintating to consider the cardinal $\aleph_\omega$ in comparison with the smaller infinite cardinals $\aleph_n$ for finite $n$. The peculiar thing to notice is that $\aleph_\omega$ is the supremum of the $\aleph_n$, and this make $\aleph_\omega$ a countable union of sets, each of which is smaller than $\aleph_\omega$. That is, $\aleph_\omega$ is a comparatively small union of comparatively small sets. When you think about it, this is an unusual property for an infinite cardinality to have. For example, $\aleph_0$ itself doesn't have this property, since a finite union of finite sets is still finite. The cardinal $\aleph_1$ doesn't have this property, since a countable union of countable sets is still countable. Similarly, $\aleph_2$ is not a small union of small sets, since the union of $\aleph_1$ many sets of size at most $\aleph_1$ still has size $\aleph_1$. Yet, $\aleph_\omega$ is a comparatively small union of comparatively small sets.

A cardinal $\kappa$ is singular, when it can be expressed as a small union of small sets (where small here means size less than $\kappa$), and otherwise it is regular. The singular/regular distinction is a fundamental divide for infinite cardinalities, and many constructions depend on it, or more precisely on the particular cofinality of the cardinal in question.

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thanks so much, I understand very cleary. Do u advise a book to help me, when I read Kunen? –  user10806 May 13 '11 at 7:26
    
The other graduate level introductory books I recommend would be Jech's Set Theory and Kanamori's The Higher Infinite. In addition, there are a variety of undergraduate level texts. –  JDH May 13 '11 at 10:16

Unless I am mistaken Kunen uses $R(\kappa)$ to denote $V_\kappa$. So I will assume it is so and give some hints regarding the exercise.

First of all note that the notion of a cofinal function is absolute in transitive $\in$-models. So first of all we have $cf(\alpha)\leq (cf(\alpha))^{R(\kappa)}$. This is because if there is a function from an ordinal $\beta$ cofinal on $\alpha$ in $R(\kappa)$ then this function is likewise in the universe and therefore the cofinality of $\alpha$ has to be less than or equal to $\beta$. So we need the opposite direction. Assume then that we have a function $f:cf(\alpha)\to\alpha$ which is cofinal. This function will certainly be in $R(\alpha+n)$ where $n$ is a natural number, because essentially a function is a subset of $\mathcal{P}(\mathcal{P}(A\cup B))$ where $A$ is the domain of the function and $B$ is its range. Therefore since $R(\alpha+n)\subset R(\kappa)$ we have that $f\in R(\kappa)$ and thus we have that $(cf(\alpha))^{R(\kappa)}\leq cf(\alpha)$.

As for (f) you have already shown that the powerset the initial ordinals and the cofinality are absolute in $R(\kappa)$. Now what's left is to show that the beth functions are absolute. To show this you use the fact that $\kappa$ is a strong limit, and like in the above paragraph use the fact that the functions between the powerset and its cardinality is inside $R(\kappa)$. Composing these facts gives that a cardinal is strongly inaccessible in $R(\kappa)$ exactly when it is really strongly inaccessible.

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You are not mistaken in regards to $R(\kappa)$, this denotes all sets of rank $<\kappa$, which is exactly $V_\kappa$. –  Asaf Karagila May 12 '11 at 17:47
    
in the second paragraph, I dont understand for opposite diraction. you take a function from cf(alpha) to alpha? –  user10806 May 17 '11 at 14:52
    
@user: The cofinality of $\alpha$ is the least ordinal $cf(\alpha)$ such that there exists some function $h:cf(\alpha)\to\alpha$ that is cofinal. Now let $\beta$ be the real cofinality of $\alpha$. This means that there exists some function $f:\beta\to\alpha$ that is cofinal. (cont.) –  Apostolos May 17 '11 at 15:18
    
@user: Now the cofinality of $\alpha$ inside $R(\kappa)$ is the least ordinal $(cf(\alpha))^{R(\kappa)}$ such that there exists a function inside $R(\kappa)$, $g:(cf(\alpha))^{R(\kappa)}\to\alpha$ that is cofinal. If we show that the aforementioned $f$ is inside $R(\kappa)$ that means that there is a cofinal function from $cf(\alpha)$ to $\alpha$ and thus, because the cofinality is defined as the least such ordinal, $(cf(\alpha))^{R(\kappa)}$ is less than $cf(\alpha)$. –  Apostolos May 17 '11 at 15:21
    
thanks it is more clear. –  user10806 May 18 '11 at 11:56

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