Two versions of "absolutely ccc"

Question

I have recently been slogging my way through Shelah's "Large continuum, oracles". Essentially from the start there has been a question needling me which I cannot seem to answer.

• In the paper, Shelah says that a forcing notion $\mathcal{P}$ is absolutely ccc if it remains ccc after forcing with any ccc notion.
• Elsewhere, I have seen it defined that a forcing notion $\mathcal{P}$ is absolutely ccc if it remains ccc after any forcing. (This would be indestructibly ccc from Bartoszyński-Judah.)

Any forcing having the Knaster property is absolutely ccc (in the strong sense), and MA$_{\aleph_1}$ implies that all ccc forcings have the Knaster property. Thus, it is consistent that the two are equivalent.

Do these two versions of absolute ccc-ness provably coincide?

Answer 1

Bartoszyński-Judah's definition is NOT what you say, it is in fact the same as Shelah's notion. In Set Theory: On the Structure of the Real Line, they define "indestructibly c.c.c." on page 177:

"Recall that a forcing notion $\mathcal{P}$ is indestructibly ccc if for every forcing notion $\mathcal{Q}$ satisfying ccc, $V^\mathcal{Q} \vDash$ "$\mathcal{P}^V$ satisfies ccc."

Then they prove (theorem 3.5.26) that if $\mathcal{P}$ has the Knaster property, then it is indestructibly ccc, per this definition. (One can also show that having the Knaster property is itself indestructible by ccc forcing.)

I have not seen elsewhere the claim that the Knaster property implies that the ccc is indestructible by any forcing. Do you have a proof of this?

Tangentially, while I was looking in Bartoszyński and Judah's book, I found a notable mistake. They claim (lemma 1.5.14) that $\mathcal{P} * \dot{\mathcal{Q}}$ has precaliber $\aleph_1$ iff $\mathcal{P}$ has precaliber $\aleph_1$ and $\Vdash_{\mathcal{P}} \dot{\mathcal{Q}}$ has precaliber $\aleph_1$. This is false. As noted in the comments on the original post, MA$_{\aleph_1}$ implies the ccc is equivalent to having precaliber $\aleph_1$. So assume MA$_{\aleph_1}$, and let $\mathcal{C}$ be Cohen forcing, and $\dot{\mathcal{T}}$ be a $\mathcal{C}$-name for a Suslin tree (by Shelah). So in $V$, the iteration $\mathcal{C} * \dot{\mathcal{T}}$ has precaliber $\aleph_1$, but $\Vdash_{\mathcal{P}} \dot{\mathcal{T}}$ does not have precaliber $\aleph_1$.

Answer 2

(The following may be nonsense, since I am not speaking from my own knowledge, I am just transcribing a theorem I found here. There may be some silly mistake, such as a "not" that I have overlooked...)

[EDIT: As Joel Hamkins has pointed out, I have indeed overlooked a crucial detail. What I wrote works only as long as $(A,B)$ is an $(\omega_1,\omega_1)$-pregap. But $S(A,B)$ may collapse $\omega_1$, in which case $(A,B)$ becomes a countable pregap. (Then $(A,B)$ is of course filled, but $F(A,B)$ trivially has the ccc as it is countable.)]

Let $(A,B)$ be a pregap in $2^\omega$. There is an absolutely defined forcing notion $S(A,B)$ which forces a separation of $A,B$ (i.e., fills the gap), and there is another absolutely defined forcing notion $F(A,B)$ which forces $(A,B)$ to be indestructible, i.e.,unfillable by a cardinal-preserving forcing.

It is known that $F(A,B)$ has the ccc iff $(A,B)$ is a gap. [EDIT: only true for $(\omega_1,\omega_1)$-pregaps.]

Now assume that $(A,B)$ is a gap. Then $F(A,B)$ has the ccc. Assume moreover that $(A,B)$ is indestructible. Then $S(A,B)$ does not have the ccc.

In the extension by $S$, the gap is now filled, hence $F(A,B)$ has lost the ccc. [EDIT: Not true if $\omega_1$ is collapsed.]

But the gap was indestructible, so in any ccc extension, the gap is still a gap, so in any ccc extension, $F(A,B)$ still has the ccc.

So the ccc-ness of $F(A,B)$ cannot be destroyed by ccc forcing, but it can be destroyed by $S$.