Power sets of union (intersection) and union (intersection) of power sets Let $U$ be the universal set, $A,B\subseteq U$. 
Corrected due to my mistakes detected in comments. 
This works only for finite sets. 
Prove these statements:
$\mathcal P(A\cup B)\;\supseteq\; \mathcal P(A)\cup \mathcal P(B)$
$\mathcal P(A\cap B)\;=\; \mathcal P(A)\cap \mathcal P(B)$
explanation by cardinality:
$$|\mathcal P(A\cup B)|\;=\;2\;^{|A|\;+\;|B|\;-\;|A\cap B|}$$
$$\;\;\;\;|\mathcal P(A)\cup \mathcal P(B)|\;=\;2\;^{|A|}+2\;^{|B|}-2\;^{|A\cap B|}$$
$$|\mathcal P(A\cup B)|\;\geq\;|\mathcal P(A)\cup \mathcal P(B)|$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;|\mathcal P(A\cap B)|\;=\;|\mathcal P(A)\cap \mathcal P(B)|\;=\;2\;^{|A\cap B|}$$
Is this correct?
 A: Your proof isn't correct : if $X$ is an infinite set, you can always find a proper subset of $X$ which has the same cardinality as $X$.  
All you need to use is the definition of the power set : $$X \in \mathcal{P}(S) \Longleftrightarrow X \subseteq S$$
proof of $\mathcal P(A\cup B)\;\supseteq\; \mathcal P(A)\cup \mathcal P(B)$ :

Let $X \in \mathcal P(A)\cup \mathcal P(B)$,
  If $X \in
> \mathcal{P}(A)$, then $X \subseteq A$, hence $X \subseteq A \cup B$
  whence $X \in \mathcal{P}(A\cup B)$.
  Likewise, if $X\in
> \mathcal{P}(B)$, then $X \in \mathcal{P}(A\cup B)$.

proof of $\mathcal P(A\cap B)\;=\; \mathcal P(A)\cap \mathcal P(B)$:  

  
*
  
*"$\subseteq$" : Let $X\in \mathcal{P}(A\cap B)$ ie $X \subseteq A \cap B$. Then $X\subseteq A$ (ie $X \in \mathcal{P}(A)$) and
  $X\subseteq B$ (ie $X \in \mathcal{P}(B)$). Hence, $X \in
 \mathcal{P}(A) \cap \mathcal{P}(B)$.
  
*"$\supseteq$" : Let $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$, that is to say $X \in \mathcal{P}(A)$ (ie $X \subseteq A$) and $X \in
 \mathcal{P}(B)$ (ie $X \subseteq B$). We hence have $X\subseteq A \cap
 B$, in other words $X \in \mathcal{P}(A\cap B)$.
  

