TITLE: Proving $f\left(\bigcup\limits_{i \in I} A_i\right) = \bigcup\limits_{i \in I} f(A_i)$.
QUESTION [5 upvotes]: I am trying to prove that Proving $f\left(\bigcup\limits_{i \in I} A_i\right) = \bigcup\limits_{i \in I} f(A_i)$. Here is my attempt.

Given $y \in Y$, we have:
\begin{align*}
y \in f\left(\bigcup\limits_{i \in I} A_i\right) & \iff \exists a \in \bigcup\limits_{i \in I} A_i, \; y = f(a) \\
& \iff \exists i \in I, \; a \in A_i, \; y = f(a) \\
& \iff \exists i \in I, \; y \in f(A_i) \\
& \iff y \in \bigcup\limits_{i \in I} f(A_i).
\end{align*}

How does this look?

REPLY [2 votes]: Let me write down little additional specification based on, that $(\exists x \in X)R(x)\Leftrightarrow (\exists x)(x \in X\land R(x))$:
$$\begin{align*}
&\exists a \in \bigcup\limits_{i \in I} A_i, \; y = f(a) \\
& \iff (\exists a)\Big(a \in \bigcup\limits_{i \in I} A_i \land y = f(a)\Big)\\
&\iff (\exists a)\Big((\exists i)(i \in I \land a \in A_i) \land y = f(a)\Big)\\
&\iff (\exists a)(\exists i)\Big(i \in I \land a \in A_i \land y = f(a)\Big)\\
\end{align*}$$
Now you can reorder/move/group existential quantifiers and logical and  and get 3'd line.