# Thread: [SOLVED] Proof involing power sets

1. ## [SOLVED] Proof involing power sets

Is it okay?

For sets A,B:
Prove that P(power set)(A) U P(B) is contained in P(AUB)

I wrote:

Let S be a subset and let S belong to P(AUB)
Then S belongs to P(AUB) iff S is contained in (AUB)
and S is contained in (AUB) iff S is contained in A and S is contained in B
iff S belongs to the power set of A and the power set of B, which is equivalent to S belongs to P(A)U P(B)

Is it okay?

2. Originally Posted by zhupolongjoe
For sets A,B:
Prove that P(power set)(A) U P(B) is contained in P(AUB)
Then S belongs to P(AUB) iff S is contained in (AUB)
Almost all of what you wrote is not correct.
This is an easy proof.
First, $\displaystyle C \subseteq D\; \Rightarrow \;P(C) \subseteq P(D)$

Second, for any set $\displaystyle X$ it is true that $\displaystyle A \subseteq A \cup X$ which means $\displaystyle P\left( A \right) \subseteq P\left( {A \cup X} \right)$.

3. Let me try this then, which seems too easy.

Let X be a set and let X be contained in A and also let X be contained in B.

So P(X) is contained in P(A) And in P(B) so P(X) is contained in P(A)UP(B).

So X is contained in (AUB), then P(X) is contained in P(AUB)

Seems wrong, but I don't know.

4. Here is the complete proof.
$\displaystyle \begin{gathered} A \subseteq A \cup B\; \Rightarrow \;P(A) \subseteq P(A \cup B) \hfill \\ B \subseteq A \cup B\; \Rightarrow \;P(B) \subseteq P(A \cup B) \hfill \\ P(A) \cup P(B) \subseteq P(A \cup B) \hfill \\ \end{gathered}$

5. Wow....that's so incredibly simple that I would never have thought of it. Thank you.

If you have some time, would you mind checking the proof I have here at the last post as well:

http://www.mathhelpforum.com/math-he...low-level.html

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# p(aub)=p(a)up(b)

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