Power set of Cartesian product I'm trying to see the differences between a Power set of a cartisian product and the cartisian product of two power sets. I have these 2 examples:
$P(\{1, 2\} \times \{3, 4\})$
$P(\{1, 2\}) \times P(\{3, 4\})$
With the first one I can arrive here, but I'm not sure how to continue to compute the power set.
$P(\{(1,3), (1,4), (2,3), (2,4)\})$
The second one I'm stuck on the cartisian product between the to result of the power set.
$\{\{\}, \{1\}, \{2\}, \{1,2\}\} \times \{\{\}, \{3\}, \{4\}, \{3,4\}\} $
How can I keep going from there? And the notation is correct?
 A: Well, on a practical and obvious level .  $A \times B$ = a set of ordered pairs so $P(A\times B)$ = a set of sets of ordered pairs.  $P(A)$ = a set of set of elements.  So $P(A) \times P(B)$ = a set of ordered pairs of sets.  
It might seem abstract but a set of ordered pairs of sets = {({..},{....})}, is completely different than a a set of sets of ordered pairs.  {{(x,y)}}.
e.g.
P({1,2} X {3,4}) = P({(1,3),(1,4),(2,3),(2,4)}) = {$\emptyset$, {(1,3)},{(1,4)},{(2,3)},{(2,4)},{(1,3),(1,4)},{(1,3),(2,3)},{(1,3),(2,4)},{(1,4),(2,3)}{(1,4),(2,4)},{(2,3),(2,4)},{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)},{(1,3),(1,4),(2,3),(2,4)}}
wherease P({1,2})X P({3,4}) = {$\emptyset$, {1},{2},{1,2})X ($\emptyset$, {3},{4},{3,4}) =
= {($\emptyset, \emptyset$),($\emptyset$, {3}), ($\emptyset$, {4}),($\emptyset$, {3,4}),({1}, $\emptyset$),({1}, {3}), ({1}, {4}),({1}, {3,4}),({2}, $\emptyset$),({2}, {3}), ({2}, {4}),({2}, {3,4})({1,2}, $\emptyset$),({1,2}, {3}), ({1,2}, {4}),({1,2}, {3,4})}
Different things. By coincidence they both have 16 elements.
In general:
$|A \times B| = |A| * |B|$
$|P(A)| = 2^{|A|}$.
So $|P(A \times B)| = 2^{|A||B|}$ while $|(P(A) \times P(B)| = 2^{|A|}2^{|B|}=2^{|A| + |B|} \ne 2^{|A||B|}$
so we know this can't be true.
A: The element $(\{1,2\},\{3\})$ is in $\mathcal{P}(\{1,2\})\times\mathcal{P}(\{3,4\})$, but it is not an element of $\mathcal{P}(\{1,2\}\times\{3,4\})$. Elements of that latter set are sets of ordered pairs, not ordered pairs of sets.
A: For finite sets they are definitely equal. $A \cap \phi = \phi$. 
