Understanding the validity of this proof for interesection of power sets Link: Let $A, B$ be sets. Show that $\mathcal P(A ∩ B) = \mathcal P(A) ∩ \mathcal P(B)$. 
In the first answer, particularly in this line:
"Let $Y \in P(A)\cap P(B)$.  
Then $Y \in P(A) $ and $Y \in P(B)$. Therefore each element of $Y$ is an element of $A$ and $B$."  
I don't see why this implies that $Y$ is an element of $A$, $B$. Clearly from an example I don't see it to be true either.  
Let $A = \{0,1\}$ and $B = \{1,2\}$.
$P(A) = \{ \{0\},\{1\},\{0,1\},\{\}\}$ and $P(B) = \{\{1\},\{2\},\{1,2\},\{\}\}$  
There isn't an element in $P(A)$ and $P(B)$ that is in $A$ or $B$, since the elements of the power sets are sets themselves, yet the elements in $A$ and $B$ aren't.
For example, let $Y = \{1\}$.
Am I misunderstanding this?
 A: The answer is not stating that $Y$ is an element of $A$ and $B$, it is saying that every element $y \in Y$ is an element of $A$ and $B$. In your example $1 \in A$ and $1 \in B$.
A: Recall that a Powerset is the set of subsets. $$Y\in \mathcal P(A)\iff Y\subseteq A$$
Then too recall the definition of subset.$$Y\subseteq A {~\iff~ \forall x~(x\in Y\to x\in A) \\ ~\iff~ \forall x\in Y~(x\in A)}$$
Thus if $Y$ is an element of the powerset of $A$, then any element of $Y$ is an element of $A$, et vice versa.

So we have that:
$${\begin{array}{l|c:l}1&Y~\in~\mathcal P(A)\cap\mathcal P(B) & \text{Assume}\\2& Y\in \mathcal P(A)~\wedge~ Y\in\mathcal P(B) & \text{Definition of Union}\\ 3&Y\subseteq A~~\wedge~~ Y\subseteq B & \text{Definition of Powerset}\\ \quad 4&\forall x\in Y~(x\in A)~\wedge~\forall x\in Y~( x\in B) & \text{Definition of Subset(eq)} \\\quad 5& \forall x\in Y~(x\in A\wedge x\in B) & \text{Distribution over universal quantifier} \\\quad 6&\forall x\in Y~(x\in (A\cap B)) &\text{Definition of intersection}\\ 7& Y~\subseteq ~(A\cap B) &\text{Definition of Suset(eq)}\\ 8& Y\in \mathcal P(A\cap B) & \text{Definition of Powerset} \end{array}\\... \text{and vice versa on every step}\\ Y\in\mathcal P(A)\cap \mathcal P(B)~ \iff ~Y\in \mathcal P(A\cap B)}$$
