MHB How can we prove $P(A \cap B) = P(A) \cap P(B)$?

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To prove that $P(A \cap B) = P(A) \cap P(B)$, one must demonstrate that each element of $P(A \cap B)$ is contained in $P(A) \cap P(B)$ and vice versa. This involves using the definitions of power sets and intersections. The proof starts by assuming an element $X$ belongs to $P(A \cap B)$, which implies $X \subseteq A \cap B$. From the properties of intersections, it follows that $X$ must also be a subset of both $A$ and $B$, thus belonging to $P(A)$ and $P(B)$. Ultimately, the equality holds as both sets contain the same elements.
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$P(A \cap B) = P(A) \cap P(B)$

How can we prove this to be true?
 
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Hi tmt,

To show two sets $X$ and $Y$ are equal, you must prove that every element of $X$ belongs to $Y$ and vice versa. In your case, you must prove that every element of $P(A\cap B)$ belongs to $P(A)\cap P(B)$, and every element of $P(A) \cap P(B)$ belongs to $P(A\cap B)$. Use the definitions of the power set and intersection $\cap$ to do it.
 
Normally, the way one proves that two sets are equal is to show they contain exactly the same elements-which is equivalent to showing they are subsets of each other.

Here's how one such proof might begin:

Suppose $X \in P(A \cap B)$. Then $X \subseteq A \cap B$.

Now $A \cap B \subseteq A$, and $A \cap B \subseteq B$, by the definition of "$\cap$".

So, $X \subseteq A$, and...
 

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