Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This problem is Exercise (F4) from Chapter VII of Kunen's text. Apparently, it is a result of Tarski. It goes as follows:

Let $ \mathbb{P} $ be a partial order. Define $ \text{c.c.}(\mathbb{P}) $ to be the least cardinal $ \theta $ such that $ \mathbb{P} $ has the $ \theta $-chain condition (i.e., every antichain in $ \mathbb{P} $ has cardinality $ < \theta $). Show that if $ \text{c.c.}(\mathbb{P}) \geq \omega $, then $ \text{c.c.}(\mathbb{P}) $ is a regular cardinal.

I was thinking, in order to solve this problem, do we need to consider a maximal incompatible family of atoms of $ \mathbb{P} $?

share|improve this question
1  
Asaf and Arthur Fischer, you guys rule this domain, so hopefully you can provide some insights. Thanks! :) –  Haskell Curry Nov 9 '12 at 19:46
    
Hehe.${}{}{}{}$ –  Asaf Karagila Nov 9 '12 at 20:03
    
Do you mean antichain in the same of comparability or compatability? –  Asaf Karagila Nov 9 '12 at 20:03
    
I have to admit that I'm not sure how to prove this, and unfortunately I don't have too much free time this evening to sit through this. However what I would probably try to do is to consider the case where this is true, let $\lambda=\operatorname{cf}(\kappa)<\kappa$ be $\mathrm{c.c.}(\Bbb P)$, so there are antichain chains of size $\kappa_\alpha$ for $\alpha<\lambda$. First show that there is a sequence of maximal antichains whose cardinality increases; then use recursion to Frankenstein an antichain of size $\kappa$ by taking a small portion from each maximal antichain. Contradiction! –  Asaf Karagila Nov 9 '12 at 20:14
1  
@Asaf: That is indeed the basic idea. –  Brian M. Scott Nov 9 '12 at 20:44

1 Answer 1

up vote 3 down vote accepted

I couldn’t figure out a way to give small hints, so I’m afraid that I’ve actually done most of it.

Let $\kappa=\operatorname{c.c.}(\Bbb P)$, and let $\lambda=\operatorname{cf}\kappa$. For $p\in\Bbb P$ let $$\alpha(p)=\sup\{|A|:A\subseteq{\downarrow\! p}\text{ and }A\text{ is an antichain}\}\;$$

Suppose that $\lambda<\kappa$, and let $A=\{p\in\Bbb P:\forall q\le p(\alpha(q)=\kappa\}$. Suppose first that $A\ne\varnothing$, and let $p\in A$. Since $\alpha(p)=\kappa>\lambda$, there is an antichain $C=\{q_\xi:\xi<\lambda\}\subseteq{\downarrow\! p}$ of cardinality $\lambda$. Let $\langle\eta_\xi:\xi<\lambda\rangle$ be a sequence of cardinals cofinal in $\kappa$, and for $\xi<\lambda$ let $C_\xi$ be an antichain in ${\downarrow\! q}$ of cardinality $\eta_\xi$; use these antichains to derive a contradiction and conclude that $A=\varnothing$.

Let $D=\{p\in\Bbb P:\alpha(p)<\kappa\}$; we’ve just shown that $D$ is dense in $\Bbb P$. Let $M$ be a maximal subset of $D$ that is an antichain in $\Bbb P$; clearly $M$ is a maximal antichain in $\Bbb P$, since $D$ is dense, so $|M|<\kappa$. Suppose that $|M|<\mu<\kappa$; since $\mu<\kappa$, there is an antichain $C$ of power $\mu$, and since $\mu>|M|$, there is a $p\in M$ compatible with $\mu$ members of $C$. Let $C_p$ be the set of members of $C$ that are compatible with $p$, and for each $q\in C_p$ fix $r_q\in({\downarrow\! q})\cap({\downarrow\! p})$; clearly $\{r_q:q\in C_p\}$ is an antichain of power $\mu$ in ${\downarrow\! p}$. Thus, $\sup\{\alpha(p):p\in M\}=\kappa$, and we can recursively construct a $\lambda$-sequence $\langle p_\xi:\xi<\lambda\rangle$ in $M$ such that $\sup\{\alpha(p_\xi):\xi<\lambda\}=\kappa$ and $\alpha(p_\eta)>\sup\{\alpha(p_\xi):\xi<\eta\}$ for each $\eta<\lambda$. Now choose suitable antichains in ${\downarrow\! p_\xi}$ for $\xi<\lambda$ and piece them together to get too big an antichain in $\Bbb P$.

share|improve this answer
    
Thank you! It seems that I was on the right track. –  Haskell Curry Nov 10 '12 at 7:56
    
@Haskell: You’re welcome! –  Brian M. Scott Nov 10 '12 at 7:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.