To prove $f(A\cap B)=f(A)\cap f(B) \Longleftrightarrow \text{f is injective}$
Beginning:
$f(A\cap B)=\{f(x): x\in (A\cap B)\}=\{f(x): x\in A ∧ x\in B\}$ Is that correct and how can I proceed?
Asked
Active
Viewed 427 times
0
-
Start from writing the definitions. Also note that $f(A\cap B)\subseteq f(A)\cap f(B)$ is always true. – Mark Apr 23 '19 at 21:41
-
How do you know that $f(A\cap B)\subseteq f(A)\cap f(B)$ is always true? – Apr 23 '19 at 22:07
-
This is very easy to prove. Let $y\in f(A\cap B)$. Then there is $x\in A\cap B$ such that $y=f(x)$. Can you see why $y\in f(A)\cap f(B)$? – Mark Apr 23 '19 at 22:09
-
I can't actually. – Apr 23 '19 at 22:09
-
If $x\in A\cap B$ then $x\in A$. And this implies $y=f(x)\in f(A)$. Also, $x\in B$ and hence $y=f(x)\in f(B)$. So $y$ is in both sets, hence it is in the intersection $f(A)\cap f(B)$. – Mark Apr 23 '19 at 22:10
-
I see, thank you. – Apr 23 '19 at 22:11