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)?

  • $\begingroup$ You're missing/ignoring the requirement that the forcing notions must be separative. $\endgroup$ – Not Mike Jan 3 '19 at 17:27
  • 1
    $\begingroup$ @NotMike I'm very new to this, and I don't understand why is that required in any sense for any of my statements/questions $\endgroup$ – Emilio Martinez Jan 3 '19 at 17:34
  • 2
    $\begingroup$ Replacing "ccc" by uncountable cardinals is a very difficult problem. Recently James Cummings, Mirna Dzamonja, and Itay Neeman have proposed a nice generalization. You can find it on arXiv. Other than this, you'd venture well into generalized properness, which is depressingly convoluted and hard to understand, main works are by Shelah and Roslanowski. $\endgroup$ – Asaf Karagila Jan 3 '19 at 17:34
  • 2
    $\begingroup$ I mean that there are a lot inherent difficulties in forcing axioms which do not revolve around $\omega$ in some significant way. $\endgroup$ – Asaf Karagila Jan 3 '19 at 17:48
  • 2
    $\begingroup$ The point is not to "just change ccc", but also to get something which is consistent and helpful in proving similar consequences to what MA provides us with. $\endgroup$ – Asaf Karagila Jan 5 '19 at 21:31

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 $q_1,q_2\le p$ 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.

  • $\begingroup$ Thanks. It was my intuition that the problem would trivilize like that, but I hadn't yet matured the subject enough to prove it. $\endgroup$ – Emilio Martinez Jan 3 '19 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.