0

How to prove $f(A) \cap f(B)=\emptyset$ for all $A,B⊂X$ with $A∩B=∅$ $\Longleftrightarrow$ $f$ is injective

For "$\Longrightarrow$"

Let $x \in A$ and $y\in B$. Since $A∩B=∅$, it is $x\neq y$ and hence $f(x)\neq f(y)$ which is the definition of injectivity.

For "$\Longleftarrow$"

No real base here...

anomaly
  • 25,364
  • 5
  • 44
  • 85

2 Answers2

0

Let $f$ be injective. Suppose $A,B\subseteq X$ and $A\cap B=\emptyset$. We want to show $f(A)\cap f(B)=\emptyset$. So assume $x\in f(A)\cap f(B)$, i.e., $x\in f(A)$ and $x\in f(B)$. Then $x=f(a)$ for some $a\in A$ and $x=f(b)$ for some $b\in B$. By injectivity, it follows that $a=b$, hence $a\in A\cap B$, contradicting $A\cap B=\emptyset$. We conclude that no $x\in f(A)\cap f(B)$ can exist, i.e., $f(A)\cap f(B)=\emptyset$, as desired

Hagen von Eitzen
  • 374,180
0

$\longrightarrow$: you cannot start with $A$ and $B$. To prove that $f$ is injective you have to start with $f(x)=f(y)$ and show that $x=y$ by coming up with suitable $A$ and $B$. Let us prove by contradiction. Suppose $x \neq y$. Take $A=\{x\},B=\{y\}$. Then $A\cap B=\emptyset$ but $f(A)\cap f(B) \neq \emptyset$. This is a contradiction so $\longrightarrow$ is proved.

$\longleftarrow$: suppose $f$ is injective and let $A$, $B$ be disjoint sets. We have to show that $f(A)\cap f(B) = \emptyset$. Once again we prove this by contradiction. Suppose $y \in f(A)\cap f(B)$. Then $y=f(a)$ for some $a \in A$ and $y=f(b)$ for some $b \in B$. But then $f(a)=f(b)$ and injectivity implies $a=b$. This gives a point in $A \cap B$, so we have a contradiction.

Kavi Rama Murthy
  • 311,013