# Image of a Union

## Homework Statement

Show that if $f: A \rightarrow B$ and $E,F \subseteq A$, then $f(E \cup F) = f(E) \cup f(F)$, and $f(E \cap F) \subseteq f(E) \cap f(F)$.

## Homework Equations

$f(E) := \{f(x)~|~ x \in E \}$.

## The Attempt at a Solution

Okay, showing $f(E \cup F) \subseteq f(E) \cup f(F)$ is rather easy. Let us look at the second direction. Let $y \in f(E) \cup f(F)$ be arbitrary. Then $y \in f(E)$ or $y \in f(F)$, which means there exists $x_1 \in E$ and $x_2 \in F$ such that $f(x_1) = y = f(x_2)$.

It isn't clear why this implies $y = f(x) \in f(E \cup F) := \{f(x) ~|~ x \in E ~or~ x \in F \}$. Certainly if $x_1 = x_2 := x$ were the case, then I could see this.

$y\in f(E)$ or $y\in f(F)$ implies that $\exists x\in E$ or $x\in F,$ that is, $x\in E\cup F,$ st. $f(x)=y.$ Hence $y\in f(E\cup F).$
Why wouldn't we have different $x$'s? For instance, consider $y=f(x) = x^2$. Given a $y$, there exist two different $x$'s.
There's no another $x$ in my statement. Just confirm if $x$ belongs to $E\cup F.$