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 $\mathbb{P}$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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