Prove that $(\bigcup\mathcal F)\setminus(\bigcup\mathcal G)\subseteq\bigcup(\mathcal F\setminus\mathcal G).$ Not a duplicate of
Suppose $F$ and $G$ are families of sets. Prove that $(\bigcup F) \setminus (\bigcup G) \subseteq \bigcup (F \setminus G)$.
This is exercise $3.4.20.a$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\mathcal F$ and $\mathcal G$ are families of sets. Prove that $(\bigcup\mathcal F)\setminus(\bigcup\mathcal G)\subseteq\bigcup(\mathcal F\setminus\mathcal G).$
Here is my proof:
Let $x$ be an arbitrary element of $(\bigcup\mathcal F)\setminus(\bigcup\mathcal G)$. This means $x\in\bigcup\mathcal F$ and $x\notin\bigcup\mathcal G$. Since $x\in\bigcup\mathcal F$, we can choose some $A_0$ such that $A_0\in \mathcal F$ and $x\in A_0$. $x\notin\bigcup\mathcal G$ is equivalent to $\forall B(B\in\mathcal G\rightarrow x\notin B)$ and in particular $A_0\in\mathcal G\rightarrow x\notin A_0$. From $A_0\in\mathcal G\rightarrow x\notin A_0$ and $x\in A_0$, $A_0\notin\mathcal G$. From $A_0\in\mathcal F$ and $A_0\notin\mathcal G$, $A_0\in\mathcal F\setminus\mathcal G$. From $A_0\in\mathcal F\setminus\mathcal G$ and $x\in A_0$, $x\in\bigcup(\mathcal F\setminus\mathcal G)$. Therefore if $x\in(\bigcup\mathcal F)\setminus(\bigcup\mathcal G)$ then $x\in\bigcup(\mathcal F\setminus\mathcal G)$. Since $x$ is arbitrary, $\forall x\Bigr(x\in(\bigcup\mathcal F)\setminus(\bigcup\mathcal G)\rightarrow x\in\bigcup(\mathcal F\setminus\mathcal G)\Bigr)$ and so $(\bigcup\mathcal F)\setminus(\bigcup\mathcal G)\subseteq\bigcup(\mathcal F\setminus\mathcal G)$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.
 A: I think your way of arguing is fine, but one can write it down more clear. Here is how I would write down your argument:
(I assume, that $\bigcup \mathcal{F}$ stands for the union of all sets in $\mathcal{F}$.)
Let $x \in \bigcup \mathcal{F} \setminus \bigcup \mathcal{G}$. That is equivalent to the condition that both $x \in \bigcup \mathcal{F}$ and $x \notin \bigcup \mathcal{G}$ hold.
Hence, there must be a set $A_0$ in the family $\mathcal{F}$ containing $x$ (because $x \in \bigcup \mathcal{F}$ holds) and no set in $\mathcal{G}$ can contain $x$ (because $x \notin \bigcup \mathcal{G}$ holds).
Therefore, no set in $\mathcal{G}$ can contain $x$ and - in particular - $A_0$ cannot be in the family $\mathcal{G}$: $A_0 \notin \mathcal{G}$.
But $A_0 \in \mathcal{F}$ and $A_0 \notin \mathcal{G}$ clearly implies $A_0 \in \mathcal{F}\setminus\mathcal{G}$, so after all,
\begin{equation}
x \in A_0 \subset \bigcup \mathcal{F}\setminus\mathcal{G}.
\end{equation}
A: Your proof seems fine, but as mentioned in the comments it could do with some tidying.
Let $\mathcal{F}=\{F_i\}_{i\in I}$ and $\mathcal{G} = \{G_j\}_{j\in J}$ be families of sets indexed by $I$ and $J$ respectively, and $x\in (\bigcup_{i} F_i)\backslash(\bigcup_{j} G_j)$. Then $x\in\bigcup_{i} F_i$ and $x\notin\bigcup_{j} G_j$. Therefore there is some $F_k\in\mathcal{F}$ such that $x\in F_k$ and $x\notin G_j$ for all $G_j\in\mathcal{G}$. Hence $F_k \in\mathcal{F}\backslash\mathcal{G}$, and so $x\in\bigcup_{t\in T}\mathcal{F}\backslash\mathcal{G}$, where $T$ indexes $\mathcal{F}\backslash\mathcal{G}$.
