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.

[tex]f(A\cap B)=f(A)\cap f(B)[/tex]

The Attempt at a Solution

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

[tex]\Rightarrow[/tex]

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

[tex]f(A\cap B)=f(A)\cap f(B)[/tex]

For: [tex]\Leftarrow[/tex]

We have to show a double inclusion so since:

[tex]A\cap B\subset A[/tex] and [tex]A\cap B\subset B[/tex] then:

[tex]f(A\cap B)\subset f(A)[/tex] and [tex]f(A\cap B)\subset f(B)[/tex]

so therefore: [tex]f(A\cap B)\subset f(A)\cap f(B)[/tex]

And the other way around:

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

[tex]f(A)\cap f(B)\subset f(A\cap B)[/tex]

So finally : [tex]f(A\cap B)=f(A)\cap f(B)[/tex]

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.

[tex]f(A\cap B)=f(A)\cap f(B)[/tex]

The Attempt at a Solution

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

[tex]\Rightarrow[/tex]

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

[tex]f(A\cap B)=f(A)\cap f(B)[/tex]

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: [tex]\Leftarrow[/tex]

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:

[tex]A\cap B\subset A[/tex] and [tex]A\cap B\subset B[/tex] then:

[tex]f(A\cap B)\subset f(A)[/tex] and [tex]f(A\cap B)\subset f(B)[/tex]

so therefore: [tex]f(A\cap B)\subset f(A)\cap f(B)[/tex]

And the other way around:

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

[tex]f(A)\cap f(B)\subset f(A\cap B)[/tex]

So finally : [tex]f(A\cap B)=f(A)\cap f(B)[/tex]

But that is given.

Hence f has to be an injective function.

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