# 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.

## Answers and Replies

What you proceeded isn't correct.
##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.##