1. ## Power Sets Proof

Prove the following:

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

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

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

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

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

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

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

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

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

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

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

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.