9
$\begingroup$

I asked this question on M.SE a while ago and got no answers, so I'm asking it here.

Let $\kappa$ be an infinite cardinal. A family $\mathcal{A}\subseteq\mathcal{P}(\kappa)$ is independent if for any $A_1,\ldots,A_n\in\mathcal{A}$ and $i_1,\ldots,i_n\in\{0,1\}$, we have $$ \left|\bigcap_{k=1}^n A_k^{i_k}\right| = \kappa $$ where $A^0 = A$ and $A^1 = \kappa\setminus A$.

Question: Is there an independent family $\mathcal{A}$ such that the Boolean algebra generated by $\mathcal{A}$, along with the subsets of $\kappa$ of size $< \kappa$, is all of $\mathcal{P}(\kappa)$?

I am particularly interested in the case $\kappa = \omega_1$, though an answer for any $\kappa$ would be interesting.

$\endgroup$

1 Answer 1

12
$\begingroup$

I think the answer is always no. Consider the algebra $\mathcal{B}$ generated by the independent family and the ideal $J$ of subsets of $\kappa$ of size $<\kappa$. Then look at $\mathcal{C} = \{ X \subseteq \kappa : \exists Y \in \mathcal{B}, X \triangle Y \in J \}$. It is easy to see that $\mathcal{C}$ is closed under set operations, and every member of $\mathcal{C}$ is obtained by a finite boolean combination of members of $\mathcal{A}$ and $J$, so it is the algebra generated by $\mathcal{A}$ and $J$. If $\mathcal{C} = \mathcal{P}(\kappa)$, then consider the homomorphism $h : \mathcal{C} \to \mathcal{P}(\kappa)/J$ given by $X \mapsto [X]_J$. Now we know $\mathcal{P}(\kappa)/J$ has antichains of size $\kappa^+$, but $\mathcal{B}$ is a free algebra and thus has the c.c.c. By the independence of $\mathcal{A}$, $\mathcal{B} \cong \mathcal{B}/J$, so $h[\mathcal{C}] \cong \mathcal{B}$, and thus $\mathcal{P}(\kappa)/J$ has the c.c.c., contradiction.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.