I think it cannot conclude that A = B because the same power set does not mean that they have the same set or same order. Am i right?Can you conclude that A = B if A and B are two sets with the same power set?

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- Sep 19th 2009, 10:39 PM #1

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- Sep 20th 2009, 02:47 AM #2

- Sep 20th 2009, 07:39 AM #3

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- Sep 20th 2009, 08:12 AM #4
I got to be honest with you: I don’t think that you have caught on to any of this material.

The above is no example at all.

Using $\displaystyle A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}$, then it is clear to anyone who understands the definitions that $\displaystyle \{1,3,5\}\in\mathcal{P}(A)\text{ BUT }\{1,3,5\} \notin\mathcal{P}(B) $.

Therefore, $\displaystyle \mathcal{P}(A)\not= \mathcal{P}(B) $ which in contrary to the given.

So start with $\displaystyle \mathcal{P}(A)= \mathcal{P}(B) $.

We are trying to show that $\displaystyle A \subseteq B\;\& \;B \subseteq A$

Let’s do it.

$\displaystyle x\in A\Rightarrow \{x\}\in\mathcal{P}(A) \Rightarrow \{x\}\in \mathcal{P}(B) \Rightarrow x\in B $. That shows that $\displaystyle A \subseteq B$.

Similarly show that $\displaystyle B \subseteq A$.

Together they mean $\displaystyle A =B $

- Sep 20th 2009, 08:57 AM #5

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Why it's not the example at all?

$\displaystyle

A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}

$

$\displaystyle

\{2,4,6\}\in\mathcal{P}(A) \text \& \{2,4,6\} \in\mathcal{P}(B)

$

Therefore: $\displaystyle

\mathcal{P}(A) = \mathcal{P}(B)

$

What is wrong with it? Please.. the more I know, I more i'm confused. yeh i'm getting lost

I understand the second part A=B. Thanks

- Sep 20th 2009, 09:15 AM #6

- Sep 20th 2009, 09:24 AM #7

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- Sep 20th 2009, 09:34 AM #8
$\displaystyle \mathcal{P}(A)= \mathcal{P}(B) $ that means that $\displaystyle C \subseteq A \Leftrightarrow C \subseteq B$

That means any subset of A is a subset of B and any subset of B is a subset of A.

C is a subset of A if and only if C is a subset of B

That means that**every**subset of A is also a subset of B and visa versa.

- Sep 20th 2009, 09:41 AM #9

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