Showing that the least cardinal $\kappa$ such that a specific poset has no strong antichain of size $\kappa$ is regular Given a poset $P$, let $\operatorname{cc}(P)$ be the least cardinal $\kappa$ such that $P$ has no strong antichains of size $\kappa$ — i.e. for any set of size $\kappa$, there exist two elements have a common lower bound. 
If $\operatorname{cc}(P)$ is infinite, is $\operatorname{cc}(P)$ always a regular cardinal?
 A: Yes, this is true. 
To first fix some notation, given an element $a$ of a poset $P$, let $P_{\leq a} = \{ x \in P : x \leq a \}$. Note that any strong antichain in the subposet $P_{\leq a}$ is also a strong antichain in $P$.
Suppose that $\kappa$ is a singular cardinal, and $\operatorname{cc} (P) \geq \kappa$. (I.e., for each $\lambda < \kappa$ there is a strong antichain in $P$ of size $\lambda$.) We will show that there is a strong antichain in $P$ of size $\kappa$.
First fix a cofinal sequence $( \lambda_\alpha : \alpha < \operatorname{cf} \kappa )$ of cardinals in $\kappa$.
There are two cases.


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*There is an $a \in P$ such that $\operatorname{cc} ( P_{\leq b} ) \geq \kappa$ for each $b \leq a$.
Fix a strong antichain $A = \{ a_\alpha : \alpha < \operatorname{cf} \kappa \}$ in $P_{\leq a}$, and for each $\alpha < \operatorname{cf} \kappa$ fix a strong antichain $A_\alpha$ in $P_{\leq a_\alpha}$ of size $\lambda_\alpha$. Note that $\bigcup_{\alpha < \operatorname{cf} \kappa} A_\alpha$ is a strong antichain in $P_{\leq a}$ (and hence in $P$) of size $\kappa$.

*For each $a \in P$ there is a $b \leq a$ with $\operatorname{cc}(P_{\leq b}) < \kappa$.
Fix a strong antichain $A$ in $P$ which is maximal with respect to the property that $\operatorname{cc}(P_{\leq a}) < \kappa$ for each $a \in A$. Note that $A$ is actually a maximal strong antichain in $P$, since given any $a \in P$ there is a $b \leq a$ with $\operatorname{cc} ( P_{\leq b} ) < \kappa$, and so $b$ (and thus also $a$) has a common lower bound with some element of $A$.
If $|A| \geq \kappa$, we're done. So assume that $|A| < \kappa$. I claim that $\sup_{a \in A} \operatorname{cc} ( P_{\leq a} ) = \kappa$. Given any $|A| \leq \lambda < \kappa$ there is a strong antichain $B$ in $P$ of size $\lambda^+$. Also, each $b \in B$ has a common lower bound with some $a \in A$. As $\lambda^+ > |A|$ is regular it follows that some $a \in A$ has a common lower bound with $\lambda^+$-many elements of $B$, and for such an $a \in A$ it follows that $\operatorname{cc} (P_{\leq a}) > \lambda^+$.
We may now recursively pick $( a_\alpha )_{\alpha < \operatorname{cf} \kappa}$ distinct elements of $A$ such that $\operatorname{cc} ( P_{\leq a_\alpha} ) > \lambda_\alpha$. If for each $\alpha < \operatorname{cf} \kappa$ we pick $A_\alpha$ a strong antichain in $P_{\leq a_\alpha}$ of size $\lambda_\alpha$, as in the first case above it follows that $\bigcup_{\alpha < \kappa} A_\alpha$ is a strong antichain in $P$ of size $\kappa$.
