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.

  • $\begingroup$ First, there is a typo in your proof: $|A|\le|X|\le\kappa$, but not $|A|\le|X|<\kappa$. $\endgroup$
    – Hanul Jeon
    Commented Mar 13, 2021 at 22:15
  • $\begingroup$ Also, $|\operatorname{dom}p\cap X|<\kappa$, not $=\kappa$. I think your proof is fine except for those points. $\endgroup$
    – Hanul Jeon
    Commented Mar 13, 2021 at 22:28
  • $\begingroup$ 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 $\endgroup$ Commented Mar 15, 2021 at 9:03


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