Finite chain condition - Variation of Martin's Axiom statement

Question

In the following $k$ and $w$ will be cardinal numbers.

Consider the classical statement $MA(k)$:

For any partial order $P$ satisfying the countable chain condition (hereafter $ccc$) and any family $D$ of dense sets in $P$ such that $|D| = k$, there is a filter $F$ on $P$ such that $F \cap d$ is non-empty for every $d$ in $D$

Let's generalize it to the statement $MA(w, k)$ that replaces the $ccc$ by any width $w$, stating:

For any partial order $P$ satisfying that every strong antichain is of cardinality $less$ than $w$ [...etc]

Eg. $MA(\aleph_1, k) = MA(k)$

Of course, $MA(w, k)$ implies $MA(w', k')$ for every $w' \geq w$ and $k' \geq k$

Now, I was wondering why $MA$ was so specific about antichains being countable, so to motivate the classical definition I tried mapping $MA$'s validity for each $w$ and $k$ pair. So far I've got:

• $MA(w, k)$ is true for all $k \leq \aleph_0$
• $MA(\aleph_1, 2^{\aleph_0}) = MA(2^{\aleph_0})$ is false, and then so it is for any $w$ and $k$ equal or greater
• $MA(\aleph_1, k) = MA(k)$ is independent from but consistent with $ZFC$ for every $\aleph_1 \leq k < 2^{\aleph_0}$
• $MA(\aleph_2, \aleph_1)$ is false, and then so it is for any $w$ and $k$ equal or greater

So $MA$ is no use stated for longer than countable antichains. But the case I can't figure out is for $w = \aleph_0$ and $k > \aleph_0$.

So the question is: Why $ccc$? What can be said about the validity of $MA$ when stated for posets sastisfying that every strong antichain is finite but given an uncountable number of dense sets? Does a filter always or never exist? Is it equivalent to the case $w = \aleph_1$ (ie. is allowing arbitrarily long but finite antichains equivalent to allowing countable ones too)?

Answer 1

The issue is that a poset where all strong antichains have finite width is trivial - at least, as far as forcing is concerned - and so the corresponding variant of Martin's Axiom is trivial too.

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To see that such forcings are trivial, the key point is the following:

Let $\mathbb{P}$ be a poset with an element $p$ such that every $q_1,q_2\le p$ have a common extension $r$ (that is, $p$ doesn't bound any nontrivial strong antichain). Then there is a $\mathbb{P}$-generic filter in the ground model already.

The proof is simple: let $G$ be the set of all conditions compatible with $p$.

Now suppose $\mathbb{P}$ is a poset where every strong antichain is finite. I claim that $\mathbb{P}$ has such a "trivializing" element $p$. For if not, we can inductively define a map $t$ from $2^{<\omega}$ to $\mathbb{P}$ such that:

• $\sigma\prec\tau\implies t(\sigma)\ge t(\tau)$.

• $t(\sigma0)\perp t(\sigma1)$.

But then the set $$\{t(0), t(10), t(110), t(1110), ...\}$$ forms an infinite strong antichain in $\mathbb{P}$.

In fact, we can do even better (since having only finite strong antichains is preserved by passing from $\mathbb{P}$ to $\mathbb{P}_{\le s}$):

If $\mathbb{P}$ has only finite strong antichains, then the set of "trivializing" $p$ is dense in $\mathbb{P}$; so every $\mathbb{P}$-generic filter is already in the ground model. (And consequently, we have "${\bf MA(\aleph_0,\infty)}$.")

So such a poset really is trivial (in the sense of forcing), not just "possibly trivial," as is the corresponding variant of Martin's Axiom.