Prove that f is injective

• Dec 4th 2012, 07:25 AM
MachinePL1993
Prove that f is injective
Hello,
I need to prove that function defined like this $\displaystyle f:X->Y$ for any two given sets X,Y is one-to-one if and only if $\displaystyle f(A \cap B)= f(A) \cap f(B)$, where $\displaystyle A,B \subseteq X$.

What do I need to do to prove this?

EDIT:

I've figured out how to prove that $\displaystyle f(A \cap B)= f(A) \cap f(B)$ when f is injective. I still do not know how to conclude that f is injective when $\displaystyle f(A \cap B)= f(A) \cap f(B)$.
• Dec 4th 2012, 07:43 AM
Plato
Re: Prove that f is injective
Quote:

Originally Posted by MachinePL1993
Hello,
I need to prove that function defined like this $\displaystyle f:X->Y$ for any two given sets X,Y is one-to-one if and only if $\displaystyle f(A \cap B)= f(A) \cap f(B)$, where $\displaystyle A,B \subseteq X$.
What do I need to do to prove this?

It is well known that for any function $\displaystyle f(A\cap B)\subseteq f(A)\cap f(B)$.

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.
• Dec 4th 2012, 07:46 AM
MachinePL1993
Re: Prove that f is injective
Now how to prove that f is injective if the above equality holds?
• Dec 4th 2012, 07:46 AM
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).
• Dec 4th 2012, 07:58 AM
MachinePL1993
Re: Prove that f is injective
Is this correct?
Let's assume indirectly that the function is not injective

Let's choose some $\displaystyle y$ such that$\displaystyle f(a)=y$ where $\displaystyle a \in A$ and $\displaystyle f(b)=y$ where $\displaystyle b \in B$, and $\displaystyle A \cap B=\emptyset$. Then $\displaystyle f(A \cap B)={\emptyset}$, because $\displaystyle A \cap B=\emptyset$, and $\displaystyle f(A) \cap f(B) = {y}$, which is a contradiction with the assumption because $\displaystyle f(A \cap B)\not= f(A) \cap f(B)$, so the function must be injective.
• Dec 4th 2012, 10:05 AM
Re: Prove that f is injective
it seems correct.
• Dec 4th 2012, 10:29 AM
Plato
Re: Prove that f is injective
Quote:

Originally Posted by MachinePL1993
Is this correct?
Let's assume indirectly that the function is not injective

That proof is a bit rough. But it is the correct idea.
If we assume that equality holds, but the function is not injective then $\displaystyle \exists \{t,s\}$ such that $\displaystyle f(t)=f(s)$ but $\displaystyle t \not= s$.

Define $\displaystyle A=X\setminus\{t\} ~\&~ B=\{t\}$.

Then $\displaystyle f(A\cap B)=\emptyset$ but $\displaystyle f(t)=f(s)\in f(A)\cap f(B)$.