To prove that $f \left( \bigcup_{i=1}^n A_i \right) = \bigcup_{i=1}^n f(A_i)$, let’s go through the proof step by step:

Proof:

Step 1: Show $f \left( \bigcup_{i=1}^n A_i \right) \subseteq \bigcup_{i=1}^n f(A_i)$

• Let $y \in f \left( \bigcup_{i=1}^n A_i \right)$. By the definition of the image of a set under $f$, this means there exists $x \in \bigcup_{i=1}^n A_i$ such that $f(x) = y$.
• Since $x \in \bigcup_{i=1}^n A_i$, there must exist some $i$ (where $1 \leq i \leq n$) such that $x \in A_i$.
• Therefore, $f(x) \in f(A_i)$, which implies $y \in \bigcup_{i=1}^n f(A_i)$.
• Thus, $f \left( \bigcup_{i=1}^n A_i \right) \subseteq \bigcup_{i=1}^n f(A_i)$.

Step 2: Show $\bigcup_{i=1}^n f(A_i) \subseteq f \left( \bigcup_{i=1}^n A_i \right)$

• Let $y \in \bigcup_{i=1}^n f(A_i)$. By the definition of union, this means there exists some $i$ (where $1 \leq i \leq n$) such that $y \in f(A_i)$.
• By the definition of the image of a set under $f$, this means there exists $x \in A_i$ such that $f(x) = y$.
• Since $x \in A_i \subseteq \bigcup_{i=1}^n A_i$, we have $x \in \bigcup_{i=1}^n A_i$.
• Therefore, $f(x) = y \in f \left( \bigcup_{i=1}^n A_i \right)$.
• Thus, $\bigcup_{i=1}^n f(A_i) \subseteq f \left( \bigcup_{i=1}^n A_i \right)$.

Step 3: Conclude the equality

• Since we have shown both $f \left( \bigcup_{i=1}^n A_i \right) \subseteq \bigcup_{i=1}^n f(A_i)$ and $\bigcup_{i=1}^n f(A_i) \subseteq f \left( \bigcup_{i=1}^n A_i \right)$, it follows that $f \left( \bigcup_{i=1}^n A_i \right) = \bigcup_{i=1}^n f(A_i).$

Would you like a deeper breakdown of any step or further examples?

Related Questions:

1. How does this property extend to infinite unions?
2. What happens if $f$ is not a well-defined function? Does the equality still hold?
3. Can this property be applied to intersections instead of unions?
4. How does the property differ if $f$ is not surjective?
5. Could this proof be rewritten using a more set-theoretic notation?

Tip:

When working with proofs involving images of sets, always use the definition of the image explicitly to construct or verify the relationship.