# Math Help - another question

1. ## another question

Can you conclude that A = B if A and B are two sets with the same power set?
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?

2. Given that $\mathcal{P}(A)= \mathcal{P}(B)$ that means that $C \subseteq A \Leftrightarrow C \subseteq B$.
Now just stop and think carefully about the implications of that statement.

3. Originally Posted by Plato
Given that $\mathcal{P}(A)= \mathcal{P}(B)$ that means that $C \subseteq A \Leftrightarrow C \subseteq B$.
Now just stop and think carefully about the implications of that statement.
A = {1,2,3,4,5,6,7,8}
B = {0,2,4,6,8,10}
C = {2,4,6}

if $C \subseteq A \Leftrightarrow C \subseteq B$, $\mathcal{P}(A)= \mathcal{P}(B)$ ?

And if A and B has the same power set. For eg:
A ={a1,a2}
B = {b1,b2}
How could A and B the same?

4. Originally Posted by zpwnchen
A = {1,2,3,4,5,6,7,8}
B = {0,2,4,6,8,10}
if $C \subseteq A \Leftrightarrow C \subseteq B$, $\mathcal{P}(A)= \mathcal{P}(B)$ ?
And if A and B has the same power set.
How could A and B the same?
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 $A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}$, then it is clear to anyone who understands the definitions that $\{1,3,5\}\in\mathcal{P}(A)\text{ BUT }\{1,3,5\} \notin\mathcal{P}(B)$.
Therefore, $\mathcal{P}(A)\not= \mathcal{P}(B)$ which in contrary to the given.

So start with $\mathcal{P}(A)= \mathcal{P}(B)$.
We are trying to show that $A \subseteq B\;\& \;B \subseteq A$
Let’s do it.
$x\in A\Rightarrow \{x\}\in\mathcal{P}(A) \Rightarrow \{x\}\in \mathcal{P}(B) \Rightarrow x\in B$. That shows that $A \subseteq B$.
Similarly show that $B \subseteq A$.
Together they mean $A =B$

5. Why it's not the example at all?
$
A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}
$

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

Therefore: $
\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

6. Originally Posted by zpwnchen
$
A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}
$

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

Therefore: $\color{red}
\mathcal{P}(A) = \mathcal{P}(B)$
Explain why you think that is true.
Because if you understand any of this, then you know it is completely false.

7. It is obvious that it's completely false.

But according to what you said here:

Originally Posted by Plato
Given that $\mathcal{P}(A)= \mathcal{P}(B)$ that means that $C \subseteq A \Leftrightarrow C \subseteq B$.
$
\{2,4,6\} \subseteq (A) \text \& \{2,4,6\} \subseteq (B)
$
Therefore, $\mathcal{P}(A)= \mathcal{P}(B)$ ?

8. $\mathcal{P}(A)= \mathcal{P}(B)$ that means that $C \subseteq A \Leftrightarrow C \subseteq B$
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.

9. $
A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}
$

Why cant we say that?
$
\{2,4,6,8\} \subseteq (A) \Leftrightarrow \{2,4,6,8\} \subseteq (B)
$
which means C is a subset of A if and only if C is a subset of B.