Mfrhtyj

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)?

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.

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 $q_1,q_2le p$ 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:

• $sigmaprectauimplies t(sigma)ge t(tau)$.




• $t(sigma0)perp t(sigma1)$.

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.