Power Sets Proof

**Jacobpm64** · September 20th 2007, 11:27 AM

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 $x \subseteq B, A \subseteq B$ .

Need some help please. Thanks.

**Plato** · September 20th 2007, 12:06 PM

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 $<br /> x \in B$ ?

**Jacobpm64** · September 20th 2007, 12:26 PM

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.

**Plato** · September 20th 2007, 12:39 PM

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 $\mathcal{P}(A) \subseteq \mathcal{P}(B), A \subseteq B$ .

**Jacobpm64** · September 20th 2007, 12:41 PM

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.

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