Trying to prove $\left(\bigcup_{\alpha}A_\alpha\right)\cap\left(\bigcup_{\beta}B_\beta\right)=\bigcup_{(\alpha,\beta)}A_\alpha\cap B_\beta$ Hi I am not sure about my approach of proving the following definition of index sets (distributivity)
$$\left(\bigcup_{\alpha}A_\alpha\right)\cap\left(\bigcup_{\beta}B_\beta\right)=\bigcup_{(\alpha,\beta)}A_\alpha\cap B_\beta.$$
I started out with
$$
\text{Let }\{A_\alpha \colon \alpha \in \mathfrak{A}\}\text{ and }\{B_\beta \colon \beta \in \mathfrak{B}\}\text{ be family of subsets of a set.}
$$
I then proceeded with
$$
\begin{aligned}
&\left(\bigcup_{\alpha}A_\alpha\right)\cap\left(\bigcup_{\beta}B_\beta\right)\\
=&\left\{ x: \exists\alpha \in\mathfrak{A} \text{ such that } x \in A_\alpha \right\} \cap\left\{x: \exists \beta  \in\mathfrak{B} \text{ such that } x  \in B_\beta \right\}  \\
=&  \left\{ x: \exists\alpha \in\mathfrak{A} \text{ such that } x \in A_\alpha \right\} \wedge  \left\{x: \exists \beta  \in\mathfrak{B} \text{ such that } x  \in B_\beta \right\}\\
=& \left\{x: \exists \alpha  \in\mathfrak{A}, \exists \beta  \in\mathfrak{B} \text{ such that }x  \in A_\alpha \wedge x  \in B_\beta \right\} \\
=& \left\{x: \exists (\alpha,\beta )  \in\mathfrak{A} \text{ such that } x  \in A_\alpha \cap B_\beta \right\}  \\
=&  \bigcup_{(\alpha,\beta)} A_\alpha \cap B_\beta.
\end{aligned}
$$
with $(\alpha,\beta)\in\mathfrak{A}\times\mathfrak{B}$.
Does this proof seem fine to you or is there something missing?
 A: No, it is not fine. In the equality
$$
\left(\bigcup_{\alpha}A_\alpha\right)\cap\left(\bigcup_{\beta}B_\beta\right)\\
=x \in \left\{  \exists\alpha \in\mathfrak{A} \colon x \in A_\alpha \right\} \cap\left\{ \exists \beta  \in\mathfrak{B} \colon x  \in B_\beta \right\} $$
the left hand side is a set, the right hand side is a proposition. A set can't be equal to a proposition. Mathematical notations have a precise meaning, they are not just abbreviations for the spoken language.
Next, the expression
$$\left\{  \exists\alpha \in\mathfrak{A} \colon x \in A_\alpha \right\} $$
is ambiguous to say the least. You want to define a set, it is made of elements. Open a brace, give a name to the elements of the set you want to define. Follow by a $\mid$ or a $:$ and state their properties. Close the brace. This is called a definition of a set by extension.
$$\left\{x :  \exists\alpha \in\mathfrak{A} \hbox{ such that } x \in A_\alpha \right\} $$
Now your proof is fine. Cheers!
A: I find proofs written with this much symbolic density non-idiomatic and hard to read. Instead, I suggest you explain in words why any element of the set on the left must also be an element of the set on the right, and vice versa.

Suppose $x \in \left(\bigcup_{\alpha\in\mathfrak{A}} A_\alpha \right) \cap \left(\bigcup_{\beta\in\mathfrak{B}} B_\beta \right)$. Then $x \in \left(\bigcup_{\alpha\in\mathfrak{A}} A_\alpha \right)$ and $x \in \left(\bigcup_{\beta\in\mathfrak{B}} B_\beta \right)$. This means that there exists $\alpha \in \mathfrak{A}$ such that $x \in A_\alpha$, and $\beta \in \mathfrak{B}$ such that $x \in B_\beta$. Therefore, $x \in A_\alpha \cap B_\beta$. ...

