Prove that f is injective

Hello,

I need to prove that function defined like this for any two given sets X,Y is one-to-one if and only if , where .

What do I need to do to prove this?

EDIT:

I've figured out how to prove that when f is injective. I still do not know how to conclude that f is injective when .

Re: Prove that f is injective

Quote:

Originally Posted by

**MachinePL1993** Hello,

I need to prove that function defined like this

for any two given sets X,Y is one-to-one if and only if

, where

.

What do I need to do to prove this?

It is well known that for **any** function .

So you must show that equality holds if and only if the function is injective.

So assume it is injective and show equality holds. Then the converse.

Re: Prove that f is injective

Now how to prove that f is injective if the above equality holds?

Re: Prove that f is injective

In general, we always have f(AnB) included in f(A)nf(B), proof by contradiction:

let A,B subsets of X. Since f(AnB) =/= f(A)nf(B) there is an element (let's call it x) in AnB such that f(x) does not belong to f(A)nf(B). Because f(x) does not belong to f(A)nf(B), x is not in A or x is not in B. So we have an element x in AnB such that f(x) is not in A or not in B this is a contradiction.

Now let f be injective. let c be an element of f(A)nf(B). There exists c_a in A such that f(c_a)=c. And there exists c_b in B such that f(c_b)=c. Because f is injective c_a=c_b. So c is the image of an element in both A and B, so he belongs to f(AnB).

Re: Prove that f is injective

Is this correct?

Let's assume indirectly that the function is not injective

Let's choose some such that where and where , and . Then , because , and , which is a contradiction with the assumption because , so the function must be injective.

Re: Prove that f is injective

Re: Prove that f is injective