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Thread: Power Sets Proof

  1. #1
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    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 $\displaystyle x \subseteq B, A \subseteq B$.


    Need some help please. Thanks.
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  2. #2
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    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 $\displaystyle
    x \in B$?
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  3. #3
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    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.
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  4. #4
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    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 $\displaystyle \mathcal{P}(A) \subseteq \mathcal{P}(B), A \subseteq B$.
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  5. #5
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    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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