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Chapter 12 of Jech & Hrbacek's "Introduction to Set Theory" asks the following as exercise 5.5:

Show that $\textrm{MA}_\kappa$ is equivalent to the statement: If $(P,\leq)$ is an ordered set satisfying the countable antichain condition, then for every collection $\mathcal{C}$ of maximal antichains with $|\mathcal{C}|\leq\kappa$ there exists a directed set $G$ such that, for every $C\in \mathcal{C}$, there exists $c\in \mathcal{C}$ and $a\in G$ so that $c\leq a$.

Where $\textrm{MA}_\kappa$ is:

Martin's Axiom $\textrm{MA}_\kappa$ $\quad$ If $(P,\leq)$ is an ordered set satisfying the countable antichain condition, then, for every collection $\mathcal{C}$ of cofinal subsets of $P$ with $|\mathcal{C}|\leq\kappa$, there exists a $\mathcal{C}$-generic set.

I can show that $\textrm{MA}_\kappa$ implies the statement of the exercise as follows:

For every maximal antichain $C\in\mathcal{C}$ define $T_C=\{p\in P\space | \space p \notin C \text{ and }\exists c \in C \text{ such that }c<p \}$. Now consider the collection $\mathcal{C}^*=\{C^*\space|\space C^*=C \cup T_C \text{ for some } C\in \mathcal{C}\}$. It is easy to verify that $\mathcal{C}^*$ is a collection of cofinal subsets satisfying the conditions of $\textrm{MA}_\kappa$, and that the $\mathcal{C}^*$-generic set is the directed set $G$ in the statement of the exercise.

I fail, however, to see how the reverse implication is true. I am assuming that we must find maximal antichains, each contained in some cofinal subset $C\in \mathcal{C}$. But, I have not yet been successful in that. What am I missing?

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It seems you're using "$p\ge q$" to denote "$p$ is at least as strong as $q$" - note that this conflicts with the more common usage (as e.g. in Kunen's book and Jech's solo book). This won't matter below, since I'll avoid using the symbol in the forcing context altogether, but it's worth mentioning.


Your intuition is exactly right - you're going to find, for each cofinal set $C$, a maximal antichain $A_C$ with $A_C\subseteq C$. Then a directed $G$ meeting each $A_C$ will yield a $\mathcal{C}$-generic filter straightforwardly.

Now anytime you hear the word "maximal," you should automatically think of Zorn's lemma. This is no exception, and there's a natural poset that leaps to mind:

Construction: Given $X\subseteq\mathbb{P}$, let $Ant(X)$ be the set of $A\subseteq X$ such that $A$ is an antichain in $\mathbb{P}$; we order $Ant(X)$ by inclusion.

It's easy to see that $Ant(X)$ always satisfies the hypotheses of Zorn's lemma, and so we get:

Fact: For any $X\subseteq \mathbb{P}$, there is an $A\subseteq X$ which is a $\mathbb{P}$-antichain which is "maximal in $X$" (that is, not strictly contained in any antichain $B\subseteq X$).

Note, however, that this does not imply that $A$ is actually a maximal antichain in the sense of $\mathbb{P}$, since $X$ could be "very small." So this is the remaining key:

Claim: If $X\subseteq\mathbb{P}$ is cofinal and $A_X\subseteq X$ is a "maximal-in-$X$" antichain, then $A_X$ is in fact a maximal antichain in the usual sense.

One you prove this, you can argue as follows: suppose $\mathbb{P}$ is c.c.c. and $\mathcal{C}$ is a collection of $\le\kappa$-many cofinal subsets of $\mathbb{P}$. For each $X\in\mathcal{C}$ let $A_X$ be a maximal-in-$X$ antichain, and let $$\mathcal{D}=\{A_X:A\in\mathcal{C}\}.$$ Clearly $\vert\mathcal{D}\vert\le\kappa$, and so we may apply our alternate version of MA$_\kappa$ to get a $G\subseteq\mathbb{P}$ such that:

  • $G$ is directed.

  • For each $A_X$ there is some $c\in G$ which is stronger than some element of $A_X$.

Now think about what happens when we look at the upwards closure $\{h\in\mathbb{P}: \exists g\in G(g\le h)\}$ of $G$ ...

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  • $\begingroup$ Thank you for your time. I am still in an introductory level, so I do not precisely understand the opening remark. My thought process was similar, but I was never able to prove the "Claim." I did not include all the details in the question lest it became boring. Here is a link to a graph that illustrates why I think the claim is probably false: [link] (imgur.com/hlQgo3E); the red dot along with A_X make an antichain in P. $\endgroup$ – Marwan Mizuri Feb 4 '19 at 18:16
  • $\begingroup$ @MarwanMizuri Remember that "antichain" refers to strong reducibility: the red dot is compatible with the rightmost green dot (the rightmost blue dot is a common extension) and in fact $A_X$ is a maximal antichain. $\endgroup$ – Noah Schweber Feb 4 '19 at 18:50

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