sqrt(-1)NeedHelp
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- Nov 17, 2013
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Hello, I am in need of some help with the following:
Let \(\displaystyle A\) and \(\displaystyle B\) be sets and let \(\displaystyle f:A\rightarrow B\). We define the function \(\displaystyle H:\mathcal P \left({A}\right)\rightarrow \mathcal P \left({B}\right)\) as follows: for \(\displaystyle C\in \mathcal P \left({A}\right)\) let \(\displaystyle H(C)\) be \(\displaystyle f(C)\).
Show: if \(\displaystyle f\) is bijective then \(\displaystyle H\) is bijective.
I know for a function to be bijective it must be injective and surjective. So this is how I am starting the proof, but I think I am going in the wrong direction with it.
Proof:
Let \(\displaystyle A\) and \(\displaystyle B\) be sets and let \(\displaystyle f:A\rightarrow B\).
Assume that \(\displaystyle f\) is bijective, that is \(\displaystyle f\) is both injective and surjective:
I am completely lost... so any guidance would help!
Let \(\displaystyle A\) and \(\displaystyle B\) be sets and let \(\displaystyle f:A\rightarrow B\). We define the function \(\displaystyle H:\mathcal P \left({A}\right)\rightarrow \mathcal P \left({B}\right)\) as follows: for \(\displaystyle C\in \mathcal P \left({A}\right)\) let \(\displaystyle H(C)\) be \(\displaystyle f(C)\).
Show: if \(\displaystyle f\) is bijective then \(\displaystyle H\) is bijective.
I know for a function to be bijective it must be injective and surjective. So this is how I am starting the proof, but I think I am going in the wrong direction with it.
Proof:
Let \(\displaystyle A\) and \(\displaystyle B\) be sets and let \(\displaystyle f:A\rightarrow B\).
Assume that \(\displaystyle f\) is bijective, that is \(\displaystyle f\) is both injective and surjective:
- Injective: \(\displaystyle \forall a,a' \in A:f(a)=f(a')\Rightarrow a=a'\)
- Surjective: \(\displaystyle \forall b \in B: \exists a \in A \ni f(a)=b\)
I am completely lost... so any guidance would help!