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.