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Problem. Prove that $\mathcal{P}(A)\cap\mathcal{P}(B)\subseteq\mathcal{P}(A\cap B)$ for all $A, B$

This is what I have so far but not sure if it's right.

Let $X \in \mathcal{P}(A \cap B)$. Then each element of $X$ is an element of $A$ and $B$, hence $X$ is also in $\mathcal{P}(A)$ and $\mathcal{P}(B)$ $\Longrightarrow$ $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$.

Now Let $Y \in \mathcal{P}(A) \cap \mathcal{P}(B)$. Then $Y \in \mathcal{P}(A)$ and $Y \in \mathcal{P}(B)$. Therefore each element of $Y$ is an element of $A$ and $B$. Hence each element of $Y$ is in $A \cap B$ $\Longrightarrow$ $Y \in \mathcal{P}(A \cap B)$. Hence we have shown that any set in $\mathcal{P}(A \cap B)$ is in $\mathcal{P}(A) \cap \mathcal{P}(B)$ and vice versa.

Sangchul Lee
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michael
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    What is $P(A)$ ? the power set of $A$ ? what have you tried ? – Surb Nov 12 '19 at 17:05
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    What have you tried? How do you usualy solve problem like this? It will help us know what kind of proof you are looking for. – Alain Remillard Nov 12 '19 at 17:06
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    If you want to show that $X\subseteq Y$, then pick an arbitrary element of $X$ and show that it follows that it must also be an element of $Y$. So... for your problem, pick an arbitrary element $x\in (\mathcal{P}(A)\cap\mathcal{P}(B))$ and use what you know about intersections and powersets to show that it follows that you must have $x\in \mathcal{P}(A\cap B)$ – JMoravitz Nov 12 '19 at 17:14
  • I added what I've tried. – michael Nov 12 '19 at 17:16
  • If all you wanted to show was that $\mathcal{P}(A)\cap \mathcal{P}(B)\subseteq \mathcal{P}(A\cap B)$ then your entire first line in your attempt is unnecessary since that is used to show the reverse. Otherwise, your attempt appears correct (though could benefit greatly from line breaks to improve readability) – JMoravitz Nov 12 '19 at 17:47

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