# $\operatorname{Add}(\kappa, \lambda)$ is $\kappa^{+}$-cc

To obtain a forcing extension $$M[G]$$ where $$2^{\kappa} = \lambda$$ we use the $$\operatorname{Add}(\kappa, \lambda)$$ poset (the set of partial functions $$\lambda \times \kappa \rightarrow 2$$ with domain size $$<\kappa$$). But in order to preserve the right cardinalities we also need to know that $$\operatorname{Add}(\kappa, \lambda)$$ is $$\kappa^{+}$$-cc.

So I would like to show in $$\mathsf{ZFC}$$ that, if $$\kappa$$ is regular and $$2^{<\kappa} = \kappa$$, then $$\operatorname{Add}(\kappa, \lambda)$$ is $$\kappa^{+}$$-cc for $$\lambda > \kappa$$ (it's trivial otherwise).

Suppose $$A$$ is a maximal antichain of $$\mathbb{P} = \operatorname{Add}(\kappa, \lambda)$$. Let $$\theta$$ be regular and large enough so that $$\lambda \times \kappa, A \in H_{\theta}$$ (so $$\mathbb{P} \in H_{\theta}$$ as well as $$H_{\theta}$$ satisfies enough of $$\mathsf{ZFC}$$).

Take $$X \prec H_{\theta}$$ such that $$\lambda \times \kappa, A \in X$$, and $$X^{<\kappa} \subseteq X$$ and $$|X| = \kappa$$. We can do this by an elementary chain argument, with an elementary chain of elementary submodels $$\langle X_i : i < \kappa\rangle$$ and at each step you add $$X_i^{<\kappa}$$ to $$X_{i+1}$$. Then take $$X = \cup X_i$$. This should work I believe using regularity of $$\kappa$$.

Now it suffices to show that $$A \subseteq X$$ so that $$|A| \leq |X| < \kappa$$. So take any $$p \in \mathbb{P}$$, and we show that $$\exists a \in A \cap X$$ so that $$a\parallel p$$.

Take $$p \in \mathbb{P}$$. Then let $$q = p \cap X$$.

Here is the point from which I start to feel uneasy with my work.

I claim that $$q = p_{|\operatorname{dom}(p) \cap X}$$ using the fact that $$\lambda \times \kappa \in X$$ and $$X^{<\kappa} \subseteq X$$- if $$\alpha \in \operatorname{dom}(p) \cap X$$, then $$p(\alpha) \in \{0, 1\} \subseteq X$$ & $$(\alpha, p(\alpha))$$ is a simple enough object so that $$(\alpha, p(\alpha)) \in X$$; it's a simple enough object since $$p(\alpha) \in X$$, $$\alpha \in X$$, so $$\{\alpha, p(\alpha)\} \in X$$ (using pairing axiom or $$X^{<\kappa} \subseteq X$$?) and this is an ordered pair, which is a sequence of length $$2$$ again and again use $$X^{<\kappa} \subseteq X$$. Kinda sketchy argument since $$X$$ isn't transitive but these are simple enough objects that I think this should be fine.

So $$q = p_{|\operatorname{dom}(p) \cap X}$$. Now since $$|\operatorname{dom}(p) \cap X|=\kappa$$ and $$\lambda \times \kappa \in X$$ and $$2 \subseteq X$$, we have that $$q \in X$$ as well as $$X^{<\kappa} \subseteq X$$ and $$q : \lambda \times \kappa \rightarrow 2$$ and $$\operatorname{dom}(q) \subseteq X$$. Again, this bothers me a bit- I'm worried about pathologies since $$X$$ isn't necessarily transitive.

Finally using $$H_{\theta} \vDash A$$ is a maximal antichain, and $$X \prec H_{\theta}$$, there is $$a \in X \cap A$$ such that $$a \parallel q$$.

We show that $$a \parallel p$$. Suppose they disagree on some input $$i$$. As $$a \in X$$ this means that $$(i, a(i)) \in X$$ as well as $$X^{<\kappa} \subseteq X$$ I think? Which again would give us that $$\{i\} \in X$$ and again $$i \in X$$. Thus $$i \in X$$ which means $$q(i) = p(i)$$ and since $$a\parallel q$$, $$a$$ can't disagree with $$p$$ on $$i$$, and so $$a\parallel p$$.

So $$|A| < \kappa$$ as desired.

I'm a bit worried about the steps where I use $$X^{<\kappa} \subseteq X$$ to extract useful elements. Would appreciate the help.

• First, there is a typo in your proof: $|A|\le|X|\le\kappa$, but not $|A|\le|X|<\kappa$. Mar 13, 2021 at 22:15
• Also, $|\operatorname{dom}p\cap X|<\kappa$, not $=\kappa$. I think your proof is fine except for those points. Mar 13, 2021 at 22:28
• Just to make sure, are you asking only about the correctness of your proof or are you also interested in other approaches as answers? The usual proof is a fairly short $\Delta$-system argument Mar 15, 2021 at 9:03