Prove that a function is surjective but not bijective

mandlao

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I'm trying to prove that the function f:P()P() defined by f(X)=X is surjective but not bijective.
In order to do this, I need to prove that f is surjective and not injective. To prove that it is surjective, I have to suppose BP() to prove that there exists AP() for which f(A)=f(B). Also, it might be easier to prove that the image of f is P(), but I don't know how to do this (I'm doing it by double inclusion but I can't seem to prove that P()X .
Thanks.
 
I'm trying to prove that the function f:P()P() defined by f(X)=X is surjective but not bijective.
In order to do this, I need to prove that f is surjective and not injective. To prove that it is surjective, I have to suppose BP() to prove that there exists AP() for which f(A)=f(B). Also, it might be easier to prove that the image of f is P(), but I don't know how to do this (I'm doing it by double inclusion but I can't seem to prove that P()X .
Clearly the function is surjective. Can you prove that?

If \(\displaystyle S=(-\infty,0)\cap\mathbb{Z}~\&~T=(-\infty,-2)\cap\mathbb{Z}\) is it true that \(\displaystyle f(S)=f(T)~?\)
 
Clearly the function is surjective. Can you prove that?

If \(\displaystyle S=(-\infty,0)\cap\mathbb{Z}~\&~T=(-\infty,-2)\cap\mathbb{Z}\) is it true that \(\displaystyle f(S)=f(T)~?\)

No, they map different sets. I'm confused, how can I prove that it is surjective?
 
If \(\displaystyle K\in\mathcal{P}(\mathbb{N})~\&~X=K\cap\mathbb{Z}\) then \(\displaystyle f(X)=~?\)

So f(X) = K. That's how you prove surjectivity in this case? (the doesn't "invert"?)
 
So f(X) = K. That's how you prove surjectivity in this case? (the doesn't "invert"?)
Do you even know the definition of surjective?

If so, what is the definition and how is it applied in this problem?
 
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