Sets and equivalence between images of sets

[mtayab1994]

Homework Statement

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)

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.

 

[LCKurtz]

Homework Statement

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)

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)

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.

For: \Leftarrow

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?

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)

But that is given.

Hence f has to be an injective function.

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