# Sets and equivalence between images of sets

1. Oct 24, 2013

### mtayab1994

1. The problem statement, all variables and given/known data

Let f be a function from E to F . Prove that f is an injective function if and only if for all A and B subsets of P(E)^2.

$$f(A\cap B)=f(A)\cap f(B)$$

3. The attempt at a solution

Well since we have "if and only if" that means we have an equivalences so for.

$$\Rightarrow$$

If f is an injective function so it's trivial to say that

$$f(A\cap B)=f(A)\cap f(B)$$

For: $$\Leftarrow$$

We have to show a double inclusion so since:

$$A\cap B\subset A$$ and $$A\cap B\subset B$$ then:

$$f(A\cap B)\subset f(A)$$ and $$f(A\cap B)\subset f(B)$$

so therefore: $$f(A\cap B)\subset f(A)\cap f(B)$$

And the other way around:

let $$y\in f(A)\cap f(B)$$ so there exists $$x\in A\cap B$$ such that f(x)=y then by the definition of an image we get that $$f(x)=y\in f(A\cap B)$$ so therefore:

$$f(A)\cap f(B)\subset f(A\cap B)$$

So finally : $$f(A\cap B)=f(A)\cap f(B)$$

Hence f has to be an injective function. Any help or any remarks would be very well appreciated.

Last edited: Oct 24, 2013
2. Oct 24, 2013

### LCKurtz

No, it may be easy but it isn't trivial. You have to prove it. Your assumption is $f$ is injective and you have to show for any $A,B$ that $f(A\cap B)=f(A)\cap f(B)$. In particular you argument must use the fact that $f$ is 1-1 somewhere.

Here you you should be assuming that for any $A,B$ that $f(A\cap B)=f(A)\cap f(B)$ and you must show that $f$ is 1-1. So what's the argument below about? Are you trying to show what you are assuming?

But that is given.

You can't just claim it. You have to prove it.