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This may be a naive question, but I'll pose it.

Is there an example of a notion of forcing $\mathbb{P}$ that has the $\kappa$-c.c. which is not also $\kappa$-Knaster Property also is not "factorable" as a product of partial orderings?

For reference, let me state the relevant definitions and a typical example.

Definition. A partial order $\mathbb{P}$ is $\theta$-c.c. if there is no antichain $A\subseteq\mathbb{P}$ of size $\kappa$.

Definition. A partial order is $\kappa$-Knaster if, for every sequence $\overline{p}=\langle p_\alpha\rangle_{\alpha<\kappa}$ of elements of $\mathbb{P}$ of size $\kappa$, there is an unbounded set $X\subset\kappa$ of ordinals such that $\langle p_\beta\rangle_{\beta\in X}\subset \overline{p}$ which is pairwise compatible.

The prototype example involves a product of Souslin trees:

If $T$ is a Souslin tree, then $T$ has $\omega_1$-c.c. and is therefore also also $\omega_1$-Knaster. However, $T\times T$ is not $\omega_1$-c.c., yet this product is $\omega_1$-Knaster.

Is this an accident of the separability of $T$ or are there "natural" (perhaps even non-separative) examples of partial orders that are not "resolvable" or "decomposable" into products (Cartesian or otherwise) of other partial orders?


After going through much confusion and re-reading my source text several times, I realize that the "prototypical example" I gave above is not $\omega_1$-Knaster. In fact, the very text I was reading from explicitly states this and for whatever reason, I did not see the word "not" there. It turns out this is the most important word in that particular sentence, so many thanks are due to Paul McKenney for raising the issue.

I think my original question was intended to be whether the Knaster property can hold for partial orders that are not decomposable into products of some sort, but obviously I got carried away trying to give some background.

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I think you mean it is $\kappa-c.c.$ but not $\kappa-$Knaster, since $\kappa-$Knaster implies $\kappa-c.c.$. –  Mohammad Golshani Feb 13 at 8:37
@MohammadGolshani: Yes, thank you!. I've edited my question. –  Everett Piper Feb 13 at 9:13
I still don't understand what you say in the second-to-last paragraph. No $\omega_1$-Aronszajn tree can be $\omega_1$-Knaster. A Souslin tree $T$ is ccc, but its square is neither ccc nor Knaster. –  Paul McKenney Feb 13 at 12:51
@Paul McKenney: Thanks for your comment. I posed the question because I had a deep misunderstanding of the concept and you alerted me to this fact. Thanks! –  Everett Piper Feb 16 at 4:49

2 Answers 2

up vote 5 down vote accepted

Let me concentrate on the case $\kappa=\omega_1.$ Then we know that $MA$ implies every $c.c.c.$ notion of forcing has the Knaster property

On the other hand the following is proved by Judah-Rosłanowski-Shelah in the paper ``Examples for Souslin forcing'':

Theorem. It is consistent that there exists a ccc Souslin forcing notion which does not satisfy the Knaster condition

Their forcing seems to satisfy the conditions you want.

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How does this answer the question about decomposition into a product? –  Monroe Eskew Feb 13 at 21:24

Jech states in Chapter 30 of his book that $\diamondsuit$ implies that there exists a Suslin tree whose boolean completion is a simple algebra. This means that it has no nontrivial subalgebras. Such a tree would be c.c.c., not Knaster, and not factorable as a product. I don't know if the construction is written up, but it might be in the book by Devlin and Johnsbraaten.

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Jech, Simple complete Boolean algebras, Israel J. Math. 18 (1974), 1-10. Jensen's forcing to add a definable real of minimal degree also works because the complete Boolean algebra he used is rigid and minimal. –  François G. Dorais Feb 13 at 22:29

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