Bijections and Power Sets - Proving

[sqrt(-1)NeedHelp]

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:

• 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!

 

[pka]

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.

Suppose that \(\displaystyle H(C) = H(D)\). That means \(\displaystyle f(C) = f(D)\).

If \(\displaystyle t\in f(C)\) then \(\displaystyle t\in f(D)\) or \(\displaystyle \left( {\exists x \in C} \right) \wedge \left( {\exists y \in D} \right)\left[ {f(x) = t = f(y)} \right]\).
What does that say about \(\displaystyle x~\&~y~\), recall that \(\displaystyle f\) is injective.

Now suppose \(\displaystyle Q\in \mathcal{P}(B)\). \(\displaystyle \left( {\forall s \in Q} \right)\left( {\exists {x_s} \in A} \right)\left[ {f({x_s}) = s} \right]\) WHY?
Let \(\displaystyle R = \left\{ {{x_s}:s \in Q} \right\}\). What is \(\displaystyle H(R)=~?\).
What does that prove?

 

[sqrt(-1)NeedHelp]

Suppose that \(\displaystyle H(C) = H(D)\). That means \(\displaystyle f(C) = f(D)\).

If \(\displaystyle t\in f(C)\) then \(\displaystyle t\in f(D)\) or \(\displaystyle \left( {\exists x \in C} \right) \wedge \left( {\exists y \in D} \right)\left[ {f(x) = t = f(y)} \right]\).
What does that say about \(\displaystyle x~\&~y~\), recall that \(\displaystyle f\) is injective.

Now suppose \(\displaystyle Q\in \mathcal{P}(B)\). \(\displaystyle \left( {\forall s \in Q} \right)\left( {\exists {x_s} \in A} \right)\left[ {f({x_s}) = s} \right]\) WHY?
Let \(\displaystyle R = \left\{ {{x_s}:s \in Q} \right\}\). What is \(\displaystyle H(R)=~?\).
What does that prove?

So since f(C)=f(D), due to being injective C=D.

As for the second part, I'm not sure I understand 100% why we say\(\displaystyle Q\in \mathcal{P}(B)\). Is it because \(\displaystyle Q\subseteq B\) (since Power sets are sets of all subsets of B)? Other than my confusion from \(\displaystyle Q\in \mathcal{P}(B)\), I see that \(\displaystyle \left( {\forall s \in Q} \right)\left( {\exists {x_s} \in A} \right)\left[ {f({x_s}) = s} \right]\) is because of f being surjective.

 

[pka]

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.

• Surjective: \(\displaystyle \forall b \in B: \exists a \in A \ni f(a)=b\)

So since f(C)=f(D), due to being injective C=D. CORRECT

As for the second part, I'm not sure I understand 100% why we say\(\displaystyle Q\in \mathcal{P}(B)\). Is it because \(\displaystyle Q\subseteq B\) (since Power sets are sets of all subsets of B)? Other than my confusion from \(\displaystyle Q\in \mathcal{P}(B)\), I see that \(\displaystyle \left( {\forall s \in Q} \right)\left( {\exists {x_s} \in A} \right)\left[ {f({x_s}) = s} \right]\) is because of f being surjective.

I quoted your OP.
You said "Surjective: \(\displaystyle \forall b \in B: \exists a \in A \ni f(a)=b\)"
Well that is exactly what I did.

If \(\displaystyle Q\in \mathcal{P}(B)\) then there is a \(\displaystyle R\in \mathcal{P}(A)\) such that \(\displaystyle H(R)=Q.\).
Go back to my reply see if you can follow my proof.