Prove that $\bigcup\mathcal F$ and $\bigcup\mathcal G$ are disjoint iff for all $A\in\mathcal F$ and $B\in\mathcal G$, $A$ and $B$ are disjoint.

Question

Not a duplicate of

Suppose $F$ and $G$ are families of sets.

Prove $\bigcup\mathcal{F}$ and $\bigcup\mathcal{G}$ are disjoint iff for all $A \in \mathcal{F}$ and $B \in \mathcal{G}$, A and B are disjoint.

This is exercise $3.4.19$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $\mathcal F$ and $\mathcal G$ are families of sets. Prove that $\bigcup\mathcal F$ and $\bigcup\mathcal G$ are disjoint iff for all $A\in\mathcal F$ and $B\in\mathcal G$, $A$ and $B$ are disjoint.

Here is my proof:

$(\rightarrow)$ Suppose $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$. Let $A$ be an arbitrary element of $\mathcal F$ and $B$ be an arbitrary element of $\mathcal G$. Let $x$ be an arbitrary element of $A$. Since $A\in\mathcal F$ and $x\in A$, $x\in\bigcup\mathcal F$. From $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$ and $x\in\bigcup\mathcal F$, $x\notin \bigcup\mathcal G$. From $x\notin \bigcup\mathcal G$ and $B\in \mathcal G$, $x\notin B$. Thus if $x\in A$ then $x\notin B$. Since $x$ is arbitrary, $\forall x(x\in A\rightarrow\ x\notin B)$ and so $A\cap B=\emptyset$. Since $A$ and $B$ are arbitrary, $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$. Therefore if $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$ then $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$.

$(\leftarrow)$ Suppose $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$. Let $x$ be an arbitrary element of $\bigcup\mathcal F$. This means $A\in\mathcal F$ and $x\in A$. Let $B$ be an arbitrary element of $\mathcal G$. From $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$ and $A\in\mathcal F$, $\forall B\in\mathcal G(A\cap B=\emptyset)$. From $\forall B\in\mathcal G(A\cap B=\emptyset)$ and $B\in\mathcal G$, $A\cap B=\emptyset$. From $A\cap B=\emptyset$ and $x\in A$, $x\notin B$. Thus if $B\in \mathcal G$ then $x\notin B$. Since $B$ is arbitrary, $\forall B(B\in\mathcal G\rightarrow x\notin B)$ and so $x\notin \bigcup\mathcal G$. Thus if $x\in\bigcup\mathcal F$ then $x\notin\bigcup\mathcal G$. Since $x$ is arbitrary, $\forall x(x\in\bigcup\mathcal F\rightarrow x\notin\bigcup\mathcal G)$ and so $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$. Therefore if $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$ then $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$.

Since $\Bigr((\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset\Bigr)$ $\rightarrow$ $\Bigr(\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)\Bigr)$ and $\Bigr(\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)\Bigr)$ $\rightarrow$ $\Bigr((\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset\Bigr)$, $\Bigr((\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset\Bigr)$ iff $\Bigr(\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)\Bigr)$. $Q.E.D.$

Is my proof valid$?$

Thanks for your attention.

Edit:

$(\leftarrow)$ Suppose $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$. Let $x$ be an arbitrary element of $\bigcup\mathcal F$. So we can choose some $A_0$ such that $A_0\in\mathcal F$ and $x\in A_0$. Let $B$ be an arbitrary element of $\mathcal G$. From $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$ and $A_0\in\mathcal F$, $\forall B\in\mathcal G(A_0\cap B=\emptyset)$. From $\forall B\in\mathcal G(A_0\cap B=\emptyset)$ and $B\in\mathcal G$, $A_0\cap B=\emptyset$. From $A_0\cap B=\emptyset$ and $x\in A_0$, $x\notin B$. Thus if $B\in \mathcal G$ then $x\notin B$. Since $B$ is arbitrary, $\forall B(B\in\mathcal G\rightarrow x\notin B)$ and so $x\notin \bigcup\mathcal G$. Thus if $x\in\bigcup\mathcal F$ then $x\notin\bigcup\mathcal G$. Since $x$ is arbitrary, $\forall x(x\in\bigcup\mathcal F\rightarrow x\notin\bigcup\mathcal G)$ and so $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$. Therefore if $\forall A\in\mathcal F\forall B\in\mathcal G(A\cap B=\emptyset)$ then $(\bigcup\mathcal F)\cap(\bigcup\mathcal G)=\emptyset$.

Answer 1

It’s just a bit more detailed than is probably necessary even at this stage, but with one small exception it’s correct. I would make one change: in the second half, after you let $x$ be an arbitrary element of $\bigcup\mathscr{F}$, you say that ‘[t]his means $A\in\mathscr{F}$ and $x\in A$’. That’s not quite right: what it actually means (and what you should say) is that there is some $A\in\mathscr{F}$ such that $x\in A$. (In fact, what you wrote couldn’t be right, simply because it makes an assertion about something that has not at that point been introduced, namely, the set $A$.)