[proofplan] We prove the equality of subsets of $B$ by extensionality. For the forward inclusion, an element in the image of the union has a preimage lying in some $S_i$, so it lies in the corresponding image $f(S_i)$. For the reverse inclusion, an element lying in some image $f(S_i)$ has a preimage in $S_i$, hence in the union. [/proofplan] [step:Show every value attained on the union is attained on one member of the family] Let $y \in f\left(\bigcup_{i \in I} S_i\right)$. By the definition of image of a subset under $f$, there exists an element \begin{align*} x \in \bigcup_{i \in I} S_i \end{align*} such that $f(x)=y$. By the definition of union of a family of sets, there exists an index $i_0 \in I$ such that $x \in S_{i_0}$. Therefore $y=f(x) \in f(S_{i_0})$. Since $f(S_{i_0}) \subset \bigcup_{i \in I} f(S_i)$, we obtain \begin{align*} y \in \bigcup_{i \in I} f(S_i). \end{align*} Thus \begin{align*} f\left(\bigcup_{i \in I} S_i\right) \subset \bigcup_{i \in I} f(S_i). \end{align*} [guided] We prove the first inclusion by taking an arbitrary element of the left-hand side and unpacking the definitions. Let \begin{align*} y \in f\left(\bigcup_{i \in I} S_i\right). \end{align*} The expression $f\left(\bigcup_{i \in I} S_i\right)$ means the set of all values $f(x)$ where $x$ belongs to $\bigcup_{i \in I} S_i$. Hence there exists \begin{align*} x \in \bigcup_{i \in I} S_i \end{align*} such that $f(x)=y$. Now we use the meaning of membership in an indexed union. The condition $x \in \bigcup_{i \in I} S_i$ means that $x$ belongs to at least one member of the family. Therefore there exists an index $i_0 \in I$ such that $x \in S_{i_0}$. Since $x \in S_{i_0}$ and $f(x)=y$, the definition of image gives \begin{align*} y \in f(S_{i_0}). \end{align*} Finally, because $f(S_{i_0})$ is one of the sets appearing in the union $\bigcup_{i \in I} f(S_i)$, membership in $f(S_{i_0})$ implies \begin{align*} y \in \bigcup_{i \in I} f(S_i). \end{align*} Since $y$ was arbitrary, this proves \begin{align*} f\left(\bigcup_{i \in I} S_i\right) \subset \bigcup_{i \in I} f(S_i). \end{align*} [/guided] [/step] [step:Show every value attained on one member is attained on the union] Let $y \in \bigcup_{i \in I} f(S_i)$. By the definition of union, there exists an index $i_0 \in I$ such that $y \in f(S_{i_0})$. By the definition of image, there exists $x \in S_{i_0}$ such that $f(x)=y$. Since $S_{i_0} \subset \bigcup_{i \in I} S_i$, we have $x \in \bigcup_{i \in I} S_i$. Hence $y=f(x)$ belongs to $f\left(\bigcup_{i \in I} S_i\right)$. Therefore \begin{align*} \bigcup_{i \in I} f(S_i) \subset f\left(\bigcup_{i \in I} S_i\right). \end{align*} [/step] [step:Conclude equality by extensionality] We have proved both inclusions \begin{align*} f\left(\bigcup_{i \in I} S_i\right) \subset \bigcup_{i \in I} f(S_i) \end{align*} and \begin{align*} \bigcup_{i \in I} f(S_i) \subset f\left(\bigcup_{i \in I} S_i\right). \end{align*} By extensionality of sets, the two subsets of $B$ are equal: \begin{align*} f\left(\bigcup_{i \in I} S_i\right) = \bigcup_{i \in I} f(S_i). \end{align*} [/step]