Express $\cup_{i\in I} P(A_i)$ in logical quantifiers .

Question

Assuming $\{A_i | i \in I\}$ is an indexed family of sets. Express $\cup_{i\in I} P(A_i)$ in logical quantifiers.

I know that $P(\cup_i\in A_i)$ is equivalent to $\forall_y\forall_x [y \in x \rightarrow \exists_i\in I(y \in A_i)]$ because$\cup_{i\in I} A_i =\{x | \exists_i\in I(x \in A_i)\}$, and we can then simply substitute and apply the definition of power set on $P(\{x | \exists_i\in I(x \in A_i)\})$ to obtain $\forall_y\forall_x [y \in x \rightarrow \exists_i\in I(y \in A_i)]$ .

However, I have been having issues translating $\cup_{i\in I} P(A_i)$into quantifiers because the Union of the family set is now on the outside of the powerset.

My attempt was to first apply the power set definition to obtain: $\cup_{i\in I}\forall_y(y \in x \rightarrow y \in A_i) $. In colloquial English, I think that it's supposed to translate as the union of all the subsets of each set from the family set $\{A_i | i \in I\}$.

Answer 1

You have only "to switch" the quantifiers ...

$x \in \mathcal P(\cup_{i \in I}A_i) \equiv x \subseteq (\cup_{i \in I}A_i) \equiv ∀y[y∈x→∃i∈I(y \in A_i)]$.

Thus :

$x \in (\cup_{i \in I}\mathcal P(A_i)) \equiv ∃i∈I(x \in \mathcal P(A_i)) \equiv ∃i∈I(x \subseteq A_i) \equiv ∃i∈I \ [∀y(y \in x \to y \in A_i)]$.