3
$\begingroup$

Given an ordinal $\alpha$ let $\operatorname{Fn}(\omega, \alpha)$ be the set finite partial functions from $\omega$ to $\alpha$. Given a cardinal $\kappa$ let $${\prod_{\alpha<\kappa}}^{\text{fin}}\operatorname{Fn}(\omega,\alpha) = \{\vec{p}\in {\textstyle \prod_{\alpha<\kappa}}Fn(\omega,\alpha): |\{\alpha<\kappa: p_\alpha \not = \emptyset\}|<\omega\}.$$

We can then turn this into a partial order by letting $\vec{p}\leq \vec{q}$ when $p_\alpha \supseteq q_\alpha$ for all $\alpha<\kappa$. Let $\mathbb{P}$ denote this partial order.

Exercise III.3.96 (p.195) of Kunen's Set Theory (2011 edition) claims that if $\kappa$ is weakly inaccessible, then $\mathbb{P}$ is $\kappa$-cc.

Question: Is weak inaccessibility necessary? More precisely, can we prove that if $\kappa$ is regular, then $\mathbb{P}$ is $\kappa$-cc?

I have something like the following proof in mind. Suppose $X\subseteq \mathbb{P}$ has size $\kappa$, and let $Y = \{y: \exists \vec{p}\in X(y = \{\alpha<\kappa: p_\alpha \not = \emptyset\})\}$. Then $Y$ is a set of finite sets and has size $\kappa$, and thus by the $\Delta$-system lemma there is some $Y'\subseteq Y$ also of size $\kappa$ with a finite kernal $r$. Let $X'$ be the corresponding subset of $X$. Then since there are $\kappa$-many elements of $X'$, $\kappa$-many of them will agree on $r$. Thus there will be $\kappa$-many which are compatible in $\mathbb{P}$.

$\endgroup$
  • $\begingroup$ Sorry, I was not thinking correctly. $\endgroup$ – William Oct 17 '15 at 9:45
  • $\begingroup$ @William No worries! $\endgroup$ – GME Oct 17 '15 at 9:46
  • $\begingroup$ I think it is true that for any cardinal $\kappa$, your forcing above is $\kappa$-c.c. $\endgroup$ – William Oct 17 '15 at 10:22
  • $\begingroup$ Is this the new Kunen or the old one? $\endgroup$ – Asaf Karagila Oct 17 '15 at 15:17
  • $\begingroup$ @AsafKaragila new. $\endgroup$ – GME Oct 17 '15 at 15:21
3
$\begingroup$

It is known that finite support iteration of $\kappa$-c.c. forcings is $\kappa$-c.c.

Fix any cardinal $\kappa$. For any $\gamma < \kappa$, $Fn(\omega, \gamma)$ has the $\kappa$-c.c.

Since $Fn(\omega, \gamma)$ consists of finite partial functions, $Fn(\omega, \gamma)$ is the same set in all forcing extensions.

Therefore the finite support product (as you have above) is the same as the finite support iteration.

So $\prod_{\gamma < \kappa}^\text{fin} Fn(\omega, \gamma)$ satisfies the $\kappa$-c.c.


For a proof without use of iterated forcing:

The following is in Jech's "Set Theory" Theorem 15.17 (ii):

Let $P_i$ be forcings of size less than $\delta$, then the finite support product of $P_i$ satisfies the $\delta^+$-c.c.

Now suppose $\kappa$ is regular but not weakly inaccessible. Then $\kappa$ is sucessor cardinal. $\kappa = \delta^+$.

Then for each $\gamma < \kappa$, $|Fn(\omega, \gamma)| \leq \delta$. So by the result above, the finite support product has the $\delta^+$-c.c.

$\endgroup$
  • $\begingroup$ Thanks! I'll look at the Jech ref. later when I get a chance. $\endgroup$ – GME Oct 17 '15 at 10:37
  • $\begingroup$ So we could get the general result by combining Jech's proof for $\kappa = \lambda^+$ with Kunen's result for weakly inaccessible cardinals. But doesn't my proof just work for both cases? $\endgroup$ – GME Oct 17 '15 at 15:55

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.