Independent families versus generators

Question

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.

Answer 1

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.