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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} $?

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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
@Asaf: That is indeed the basic idea. – Brian M. Scott Nov 9 '12 at 20:44
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 $\Bbb P$.

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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

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