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We say that $\mathbb{P}$ is a complete suborder of $\mathbb{Q}$, if it is a suborder, and maximal antichains in $\mathbb{P}$ remain maximal antichains in $\mathbb{Q}$

As the title says, is every c.c.c. non-atomic partial order of size $\omega_1$ a union of countable complete suborders?

If the answer is no, what are some common c.c.c. forcings which are this way? eg. Suslin trees.

Thanks

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    $\begingroup$ math.stackexchange.com/questions/1902415/… $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 25, 2016 at 18:03
  • $\begingroup$ I never know which forum to post a question on, so after a while of no answer I thought it was maybe better suited to here, sorry. What's the usual protocol for future reference? $\endgroup$
    – Horse
    Commented Aug 25, 2016 at 18:06
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    $\begingroup$ This question is fine for MathOverflow. Some people get irritated when you post to both forums at the same time. $\endgroup$
    – Joel David Hamkins
    Commented Aug 25, 2016 at 18:17
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    $\begingroup$ Wait more than one day, a week is probably a good idea. Before 60 days old you can have it migrated by flagging for a moderator attention. AND ALWAYS add links to both posts, preferably in the post itself. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 25, 2016 at 18:22
  • $\begingroup$ That makes sense, I'll remember that for the future. Thanks for the help. $\endgroup$
    – Horse
    Commented Aug 25, 2016 at 18:28

3 Answers 3

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In response to Joel's comment, here is an argument showing that it is consistent that every ccc poset of size $\omega_1$ is the union of its nonatomic countable complete subposets.

Suppose $\text{MA}_{\omega_1}$ holds. It follows that there are no Suslin algebras, so any ccc poset must add a real. Furthermore, by a theorem of Pawlikowski, any poset of size $<\mathbf{add}(\mathcal{M})$ that adds a real must add a Cohen real. Therefore any ccc poset of size $\omega_1$ adds a Cohen real.

Now let $\mathbb{P}$ be such a poset and take $p\in\mathbb{P}$. Let $p\in A\subseteq\mathbb{P}$ be a (countable) maximal antichain. By the fact above, the cone $\mathbb{P}\upharpoonright p$ adds a Cohen real, so the usual forcing theory gives us a complete subposet $\mathbb{Q}\subseteq\mathbb{P}\upharpoonright p$ which is forcing equivalent to $\mathrm{Add}(\omega,1)$. Now, this equivalence and the fact that $\mathbb{Q}$ is ccc yield that there is a countable dense $\mathbb{Q}'\subseteq\mathbb{Q}$. Taking all of this together, $\mathbb{Q}'\cup\{p\}\cup A$ is a countable complete subposet of $\mathbb{P}$ containing $p$.

Janusz Pawlikowski, MR 1825187 Cohen reals from small forcings, J. Symbolic Logic 66 (2001), no. 1, 318--324.

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  • $\begingroup$ There is nothing very special about $\omega_1$ in this argument, but ccc seems to be crucial, both to see that $\mathbb{P}$ adds a Cohen real and that the complete subposet adding that real can be taken to be countable. Zapletal has a model where any poset of size $\omega_1$ adds a Cohen real (however the construction is specific to $\omega_1$) but I don't see how to get a small subposet adding that real in the non-ccc case. $\endgroup$
    – Miha Habič
    Commented Aug 26, 2016 at 19:50
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Noah's affirmative answer is correct with the definition that you've given, but if you want to insist that the suborder is also non-atomic, then the answer can be negative.

One easy way to see this is to note that if $\mathbb{Q}$ is a complete suborder of $\mathbb{P}$, then every forcing extension by $\mathbb{P}$ admits an intermediate forcing extension by $\mathbb{Q}$. So if $\mathbb{P}$ is the forcing notion arising from a Suslin tree, for example, then it is c.c.c., but it has no nontrivial countable complete suborders, since all such suborders are forcing equivalent to adding a Cohen real and forcing with a Suslin tree adds no reals.

More generally, one can get a concrete example in ZFC+CH. Let $\mathbb{P}$ be the forcing to add a random real, which is c.c.c., but this forcing adds no Cohen reals, and thus can have no nonatomic complete countable suborders.

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    $\begingroup$ The random real example has size continuum, rather than size $\omega_1$. So it is really a ZFC+CH example. Can anyone give a pure ZFC example? $\endgroup$
    – Joel David Hamkins
    Commented Aug 25, 2016 at 18:50
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    $\begingroup$ I think there might not be a ZFC example. In a model of MA where CH fails any ccc poset of size $\omega_1$ adds a Cohen real (this follows from a theorem of Pawlikowski). I think that you can use this to find a large variety of countable complete subposets (take any given condition, apply the above fact to the cone below it to find a countable complete subposet of the cone and enlarge it to be complete in the whole poset). $\endgroup$
    – Miha Habič
    Commented Aug 26, 2016 at 0:42
  • $\begingroup$ Very nice, Miha! Why not write that up and post an answer? $\endgroup$
    – Joel David Hamkins
    Commented Aug 26, 2016 at 1:07
  • $\begingroup$ You can place every element in a countable maximal antichain, and then do your cone idea below each element in the antichain. This will give a countable complete suborder containing the given element. So an affirmative answer is consistent with ZFC. $\endgroup$
    – Joel David Hamkins
    Commented Aug 26, 2016 at 1:27
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Suppose $\mathbb{Q}$ is c.c.c. Then every element $a\in\mathbb{Q}$ is contained in a countable maximal antichain, $A_a\subseteq\mathbb{Q}$. Each $A_a$ is a complete suborder of $\mathbb{Q}$ (the only maximal antichain in $A_a$ is $A_a$ itself, and this is a maximal antichain in $\mathbb{Q}$), and $\mathbb{Q}=\bigcup_{a\in A} A_a$.

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