1. ## Power Sets Proof

Prove the following:

Let $A$ and $B$ be sets. Then $A \subseteq B$ iff $\mathcal {P}(A) \subseteq \mathcal{P}(B)$.

Here is what I am thinking, but I'm pretty sure there is some serious problems:

Let $A$ and $B$ be sets.
First Direction: Assume $A \subseteq B$. Let $x \subseteq A$. Since $x \subseteq A, x \in \mathcal {P}(A)$. Since $A \subseteq B, x \subseteq B$. Because $x \subseteq B, x \in \mathcal{P}(B)$. Since $A \subseteq B, \mathcal {P}(A) \subseteq \mathcal{P}(B)$.

Second Direction: Assume $\mathcal{P}(A) \subseteq \mathcal{P}(B)$. Let $x \in \mathcal{P}(A)$. Since $x \in \mathcal {P}(A), x \subseteq A$. Since $\mathcal{P}(A) \subseteq \mathcal{P}(B), x \in \mathcal{P}(B)$. Since $x \in \mathcal{P}(B), x \subseteq B$. Since $x \subseteq A$ and $x \subseteq B, A \subseteq B$.

2. You have the symbolic notation confused.
If $A \subseteq B$ then prove that $\wp \left( A \right) \subseteq \wp \left( B \right)$.
Start by picking a point $G \in \wp \left( A \right)$, by definition $G \subseteq A$.
Now can you finish be concluding that $G \in \wp \left( B \right)$?

Next start with $\wp \left( A \right) \subseteq \wp \left( B \right)$ pick $x \in A$.
We know that $\left\{ x \right\} \subseteq A$. Can you finish by showing $
x \in B$
?

3. Okay, let's see what I can do here.

First way.
Assume $A \subseteq B$.
Let $G \in \mathcal{P}(A)$.
Thus $G \subseteq A$.
Since $A \subseteq B, G \subseteq B$.
Because $G \subseteq B, G \in \mathcal{P}(B)$.
Hence, if $A \subseteq B$, then $\mathcal{P}(A) \subseteq \mathcal{P}(B)$.

Second way.
Assume $\mathcal{P}(A) \subseteq \mathcal{P}(B)$.
Let $x \in A$.
Thus, {x} $\subseteq A$.
Since {x} $\subseteq A, x \in \mathcal{P}(A)$.
Since $\mathcal{P}(A) \subseteq \mathcal{P}(B), x \in \mathcal{P}(B)$.
Since $x \in \mathcal{P}(B)$, {x} $\subseteq B$.
Because {x} $\subseteq B, x \in B$.
Hence, if $\mathcal{P}(A) \subseteq \mathcal{P}(B), A \subseteq B$.

Is this okay? Or did I do something wrong again? I'm pretty confused.

4. Since {x} $\subseteq A, \color{red}{\{x\}} \in \mathcal{P}(A)$.
Since $\mathcal{P}(A) \subseteq \mathcal{P}(B), \color{red}{\{x\}} \in \mathcal{P}(B)$.
Since $\color{red}{\{x\}} \in \mathcal{P}(B)$, {x} $\subseteq B$.
Because {x} $\subseteq B, x \in B$.
Hence, if $\mathcal{P}(A) \subseteq \mathcal{P}(B), A \subseteq B$.

5. Thanks, that's what I had written on my paper.

I just screwed up when I typed it out.

You've been a great help.

Thanks so much.