Prove union/intersect of two powersets is/is not equal to powerset of their union? I am supposed to show whether or not the union/intersection of two powersets is equal to the powerset of the union/intersection, respectively. 
I found these two links that have both the proofs, and say that the union is not always equal, but the intersection is always equal. The proofs seem very similar to me. Can someone help me understand why one shows they are always equal and the other is only sometimes?
https://proofwiki.org/wiki/Union_of_Power_Sets
https://proofwiki.org/wiki/Intersection_of_Power_Sets
 A: Let's say we have two sets $S$ and $T$. Use $P(\cdot)$ to denote the power set.
The reason this statement doesn't always work for unions is because every element of $P(S) \cup P(T)$ can either only have elements of $S$ or $T$, but not both. Thus, any subset of $S \cup T$ (which is also an element of $P(S \cup T)$) that contains at least one element from both $S$ and $T$ will be absent from $P(S) \cup P(T)$, and this implies $P(S) \cup P(T) \neq P(S \cup T)$ unless there is no element that is in one but not the other (i.e. in $S$ or not $T$, or vice-versa). Ricardo Allegrone provided a nice counterexample.
This statement works for intersections since every element of $S \cap T$ is in both $S$ and $T$. Thus, any element of $e \in P(S \cap T)$ is a set of elements having the property that each of its elements in both $S$ and $T$. Thus, $e \subseteq S$ and $e \subseteq T$, meaning $e \in P(S) \cap P(T)$. From this it follows that $P(S \cap T) \subseteq P(S) \cap P(T)$. 
To prove the other direction (namely that $P(S \cap T) \supseteq P(S) \cap P(T)$), the logic is similar: any $e \in P(S \cap T)$ has the property that it is a set comprised of only elements which are in both $S$ and $T$. But then this would mean that $e \subseteq P(S)$, since all of its elements are in $S$. It would also mean that $e \subseteq P(T)$, as all of its elements are in $T$. Thus, $e \in P(S) \cap P(T)$, and these two results collectively imply that $P(S) \cap P(T) = P(S \cap T)$. 
A: The powerset of intersection is equal to the intersection of powersets, i.e. $P(A\cap B)=P(A)\cap P(B)$, can be proven as follow:
1) $P(A \cap B) \subseteq P(A) \cap P(B)$


*

*$\forall E \in P(A \cap B)$

*$[\forall x \in E(x \in A \cap B \equiv x \in A \land x \in B)]$

*$(E \subseteq A) \land (E \subseteq B)$

*$(E \in P(A)) \land (E \in P(B))$

*$E \in P(A) \cap P(B)$

*$\therefore P(A\cap B)\subseteq P(A) \cap P(B)$
2) $P(A) \cap P(B) \subseteq P(A \cap B)$


*

*$\forall E \in P(A) \cap P(B)$

*$[\forall x \in E (x \in A \land x \in B \implies x \in (A \cap B))]$

*$E \in P(A \cap B)$

*$\therefore P(A) \cap P(B) \subseteq P(A \cap B)$
Together, these give $P(A) \cap P(B) = P(A \cap B)$

To see that $P(A\cup B)\neq P(A)\cup P(B)$, simply consider the cardinality of each sides - $2^{|A|+|B|} = |P(A)\cup P(B)|\leq 2^{|A|}+2^{|B|}$ doesn't necessarily hold. 
One example would be when $A=\{1\},B=\{2\}$
$P(A\cup B)=\{\{\},\{1\},\{2\},\{1,2\}\}$
$P(A)\cup P(B)=\{\{\},\{1\},\{2\}\}$
