5

Let : $X \to Y$ be a function. Show that if $f$ is injective then $f(A \cap B) = f(A) \cap f(B)$ for sets $A \subseteq X$ and $B \subseteq X$.

My answer :

Suppose $f$ is injective and $f(x) \in f(A \cap B) \Leftrightarrow x \in A \cap B \Leftrightarrow x \in A$ and $x \in B \Leftrightarrow $ $ f(x) \in f(A)$ and $f(x) \in f(B) \Leftrightarrow f(x) \in f(A) \cap f(B)$. Therefore as each step is an equivalence it can be read backwards so $f(A \cap B) \subseteq f(A) \cap f(B)$ and $f(A) \cap f(B)$ $ \subseteq f(A \cap B)$ meaning $f(A \cap B) = f(A) \cap f(B)$.

What I don't understand is where this proof breaks down if $f$ is not injective ?

Abby
  • 145
Namch96
  • 918

2 Answers2

9

It breaks reading backwards here:

$x \in A$ and $x \in B$ $\Leftrightarrow f(x) \in f(A)$ and $f(x)\in f(B)$

You can't guarantee that it is the same $x$ that gives the value $f(x)$. To be more exact, you should have taken $y \in f(A)\cap f(B)$. Then $y \in f(A)$ and $y \in f(B)$. So exists $x_1\in A$ and $x_2 \in B$ such that $f(x_1)=f(x_2) = y$. Injectivity assures that $x_1=x_2$, and then your argument follows.

Hakim
  • 10,213
Ivo Terek
  • 77,665
4

If $f$ is not injective, there might be some $y\notin A\cap B$ such that $f(y)=f(x)$.

For example if you take $A=B=\{x\}$ you have that $\{y,x\}\subseteq\{w\mid f(w)=f(x)\}$. But of course $y\notin A\cap B=\{x\}$.

Asaf Karagila
  • 393,674