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?

  • $\begingroup$ One also sometimes sees a weaker notion, asserting only that $\cal P$ remains ccc after forcing with $\cal P$ itself. This is equivalent to saying that ${\cal P}\times{\cal P}$ is ccc. $\endgroup$ – Joel David Hamkins Mar 23 '12 at 17:32
  • $\begingroup$ Be careful! Martin's Axiom implies that any $ccc$ poset of size less than the continuum is Knaster. $\endgroup$ – saf Mar 24 '12 at 0:59
  • $\begingroup$ @saf: See e.g., Jech (3rd ed.), Theorem 16.21, p.277. The proof actually gives the slightly stronger result that MA$_{\aleph_1} $implies that all ccc posets have precalibre $\aleph_1$. $\endgroup$ – user642796 Mar 24 '12 at 8:01
  • $\begingroup$ Thanks, Arthur. I stand corrected! (indeed, if there is a non-Knaster ($ccc$) poset, then there is one of size $\aleph_1$.) The fact I had in mind is that any $ccc$ poset of size $<\mathfrak{m}$ is $\sigma$- centered. $\endgroup$ – saf Mar 24 '12 at 23:08
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    $\begingroup$ How do you prove that the Knaster property is indestructible? $\endgroup$ – Monroe Eskew Jun 8 '13 at 6:45

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 $\aleph_1$.


(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 $S$.

  • $\begingroup$ Martin, I'm a bit confused, because you are saying that F(A,B) is absolute and loses in the ccc in any forcing extension in which the gap is filled. But suppose that we force everything here to be countable? In that extension, the gap will be filled and the forcing will be ccc because it was made countable. So what you say can't be literally true. $\endgroup$ – Joel David Hamkins Mar 23 '12 at 18:29
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    $\begingroup$ It seems that the explanation is that equivalence you mention only works provided that (A,B) remains an $(\omega_1,\omega_1)$-pre-gap, which isn't itself preserved by all forcing, although it would be preserved by $\omega_1$-preserving forcing. $\endgroup$ – Joel David Hamkins Mar 23 '12 at 18:34
  • $\begingroup$ So I am unsure the example works now, because the forcing to fill the gap must collapse $\omega_1$, which destroys the hypothesis necessary to conclude that $F(A,B)$ is not ccc. $\endgroup$ – Joel David Hamkins Mar 23 '12 at 18:39
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    $\begingroup$ To further Joel's point (or confusion) I believe that the usual construction of a Hausdorff gap yields an indestructible gap. It thus seems that this answer would contradict what I've stated in the question about the two formulations of absolute ccc-ness being consistently equivalent. $\endgroup$ – user642796 Mar 24 '12 at 6:36
  • $\begingroup$ Yes, it seems you are right. I have edited the post to point the GAP in my reasoning. $\endgroup$ – Goldstern Mar 26 '12 at 6:52

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