To prove these properties of functions, we will proceed step by step. Let's break down each part of the proof.

Problem Recap:

We are given a function $f: X \to Y$, and two subsets $A$ and $B$ of $X$. We need to prove the following two properties:

I. $f(A \cup B) = f(A) \cup f(B)$

II. $f(A \cap B) \subseteq f(A) \cap f(B)$

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Proof of I: $f(A \cup B) = f(A) \cup f(B)$

To prove this equality, we need to show two things:

1. $f(A \cup B) \subseteq f(A) \cup f(B)$
2. $f(A) \cup f(B) \subseteq f(A \cup B)$

1. $f(A \cup B) \subseteq f(A) \cup f(B)$

• Let $y \in f(A \cup B)$. Then, by the definition of image, there exists some $x \in A \cup B$ such that $f(x) = y$.
• Since $x \in A \cup B$, this means that $x \in A$ or $x \in B$.
• If $x \in A$, then $f(x) \in f(A)$, and if $x \in B$, then $f(x) \in f(B)$.
• Therefore, $y = f(x) \in f(A) \cup f(B)$.

This shows that $f(A \cup B) \subseteq f(A) \cup f(B)$.

2. $f(A) \cup f(B) \subseteq f(A \cup B)$

• Let $y \in f(A) \cup f(B)$.
• This means that either $y \in f(A)$ or $y \in f(B)$.
• If $y \in f(A)$, then there exists some $x \in A$ such that $f(x) = y$.
• Similarly, if $y \in f(B)$, there exists some $x \in B$ such that $f(x) = y$.
• In either case, $x \in A \cup B$, so $f(x) = y \in f(A \cup B)$.

This shows that $f(A) \cup f(B) \subseteq f(A \cup B)$.

Since both inclusions hold, we have $f(A \cup B) = f(A) \cup f(B)$.

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Proof of II: $f(A \cap B) \subseteq f(A) \cap f(B)$

To prove this inclusion, we need to show that if $y \in f(A \cap B)$, then $y \in f(A) \cap f(B)$.

• Let $y \in f(A \cap B)$. Then, there exists some $x \in A \cap B$ such that $f(x) = y$.
• Since $x \in A \cap B$, we know that $x \in A$ and $x \in B$.
• Therefore, $f(x) \in f(A)$ and $f(x) \in f(B)$, which means $y \in f(A) \cap f(B)$.

Thus, we have $f(A \cap B) \subseteq f(A) \cap f(B)$.

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Conclusion:

We have proved the following two properties:

1. $f(A \cup B) = f(A) \cup f(B)$
2. $f(A \cap B) \subseteq f(A) \cap f(B)$

Do you want more details or have any questions about the proof?

Here are some related questions you might explore:

1. What conditions would make $f(A \cap B) = f(A) \cap f(B)$?
2. Can you provide a counterexample where $f(A \cap B) \neq f(A) \cap f(B)$?
3. How does the proof change if $f$ is injective?
4. What happens to the properties if $f$ is surjective?
5. Can these properties be extended to infinite unions or intersections?

Tip: In proving set equalities, always check for two inclusions: one subset of the other and vice versa.