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Say we have disjoint sets $A$ and $B$ where $A = \{1\}$ and $B = \{2\}$.

Thus the power set of $A$ is $\{\emptyset,1\}$ and the power set of $B$ is $\{\emptyset,2\}$.

I know the intersection set of $A$ and $B$ is $\emptyset$

My question is this: Given the above information, does that mean the intersection set between the power set of $A$ and the power set of $B$ is also the empty set?

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    $\begingroup$ How come you can't answer this yourself? You have literally all the needed information. $\endgroup$ – Asaf Karagila Oct 13 '16 at 19:45
  • $\begingroup$ Please learn MathJax. $\endgroup$ – Em. Oct 13 '16 at 20:06
  • $\begingroup$ Hey guys, my friend in another class needed some help I guess so I lent her my account so she could ask. She doesn't know how to use MathJax otherwise she probably would have. $\endgroup$ – Matthew Graham Oct 13 '16 at 20:43
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More properly, the powersets are { $\emptyset$, { 1 } } and { $\emptyset$, { 2 } }.

Their intersection is then { $\emptyset$ }, not $\emptyset$.

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No, the intersection of P(A) and P(B) is the set containing the empty set, which is itself not an empty set since it contains a set.

$\mathcal{P}(A) \cap \mathcal{P}(B) = \{\emptyset\}\neq \emptyset$

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You write:

Thus the power set of $A$ is $\{\emptyset,1\}$ and the power set of $B$ is $\{\emptyset,2\}$.

No, it isn't. In particular, $1\nsubseteq A$ and $2\nsubseteq B$. However since $1\in A$, $\{1\}\subseteq A$ and therefore $\{1\}\in\mathcal P(A)$ (where $\mathcal P(A)$ denotes the power set of $A$). Indeed, $\mathrm P(A)=\{\emptyset,\{1\}\}$. Equivalently, $\mathcal P(B)=\{\emptyset,\{2\}\}\ne\{\emptyset,2\}$.

Looking at the two sets, you can see that there is one element that is in both power sets: The empty set $\emptyset$. Therefore that is also an element of the intersection. Indeed, it's the only element that's in both sets, therefore it's the only set in the intersection.

Therefore $\mathcal P(A)\cap \mathcal P(B) = \{\emptyset\}$.

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