Give an example where equality doesn't hold in $f(C\cup D)\subseteq f(C)\cup f(D)$

Question

For $S$ in the domain of a function $f$, let $f(S)= \{f(x): x\in S\}$. Let $C$ and $D$ be subsets of the domain of $f$. Give an example where equality doesn't hold in $f(C\cup D)\subseteq f(C)\cup f(D)$.

I have proved $f(C\cup D)\subseteq f(C)\cup f(D)$ by definition, but I can't come up with an example where equality doesn't hold.

If $x\in C$ or $x\in D$, $f(x)\in f(C)$ or $f(D)$, thus if $x\in(C∪D)$, $f(X)\in f(C) \cup f(D)$?

PS: this question is from Mathematical Thinking Problem-Solving and Proofs. Second Edition.

Answer 1

What you are trying to prove is wrong. It is always true that $f$ and $\cup$ commute meaning that $f(\cup A_i)=\cup f(A_i)$. Take a look here !

Answer 2

I think you meant $f(C\cap D)\subseteq f(C)\cap f(D)$. Let me give you an example where inequality does not hold.

$\text{Let }X = \{1, 2\} \text{ and } f(1) = f(2) = 1$.

$\implies f[\{1\}\cap\{2\}]=\varnothing \text{ and }f(\{1\})\cap f[\{2\}]=\{1\}$

$\implies f[\{1\}\cap\{2\}] \neq f[\{1\}]\cap f[\{2\}]$