Set Theory, Universal Set If I have two sets A and B, given that $A := \{1, 4, 9, 16\}$ and $B := \{1, 8, 27\}$. Then is it correct to assume that the universal set is $U := \{1, 2, 3, ..., 27\}$?
Another thing, since the difference of two sets $A$ and $B$ is $A \cap B^c$, is $\mathcal{P}(A)-\mathcal{P}(B) = \mathcal{P}(A) \cap \mathcal{P}(B)^c$ or $\mathcal{P}(A) \cap \mathcal{P}(B^c)$? Thanks a lot!
 A: This depends on the context. Usually the universal set is made clear beforehand. As for the second question, the former is true. Maybe it's a bit more clear if you set $C := \mathcal{P}(A)$ and $D := \mathcal{P}(B)$. Now apply the definition of $C - D$ and afterwards, just 'remember' what $C$ and $D$ where. 
As a side note, I'd assume the universal set here is $\mathbb{N}_0$, but really there is no way to know without more information on what you're working with.
A: "Then is it correct to assume that the universal set is U = {1, 2, 3, ..., 27}?"
The only thing you can know for certain is $\{1,4,8,9,16, 27\} \subseteq U$.  But you don't need to know what the universal set is most of the time.  If unstated it probably is simply anything that is conceivable. 
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hmmm... $P(A) - P(B)$ are all the subset of $A$ that are not subsets of $B$.
That is $P(A) \cap P(B)^c$.
$P(B^c)$ and $P(B)^c$ are not the same. The former is the subsets of elements not in $B$ and the latter is everything that is not a subset of $B$. $P(B^c) \subset P(B)^c$ but $P(B)^c \not \subset P(B^c)$.  
Example: if $A = \{1,2,3\}$ and $B= \{2,3,4\}$ and let the universal set $U = \{1,2,3,4,5\}$
$P(A) - P(B) = \{\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}$ 
$P(B)^c = $ anything that isn't a subset of $B$ which if our universal set is the subsets of $U$ is everything that has $1$  or $5$ as an element.
$P(B^c) = P (\{1, 5\}) = \{\emptyset, \{1\},\{5\},\{1,5\}\}$
$P(A) \cap P(B)^c = P(A) - P(B)$ but $P(A) \cap P(B)^c =\{\emptyset, \{1\}\}$. 
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In a comment in the other answer you wrote that the original problem is "The problem is "Given the sets A={1, 4, 9, 16} and B={1, 8, 27}. Determine the sets in P(B-A) and P(B)-P(A)"
Since $B- A = \{8,27\}$ the $P(B-A)$ are the subsets of $\{8,27\}$. i.e $\{\emptyset, \{8\},\{27\}, \{8,27\}\}$.
$P(B)$ are the $2^3 = 8$ subsets of $B$.  $\{\emptyset, \{8\},\{27\}, \{8,27\}, \{1\},\{1,8\},\{1,27\}, \{1,8,27\}\}$
$P(A)$ are the subsets of $A$.  I'm not going to list them. There are $2^{4} =16$ of them.  But 
$P(B) - P(A)$ are the subsets of $B$ that are not subsets of $A$. As $B$ and $A$ have only the element $1$ in common, the only subsets that have in common are $\emptyset$ and $\{1\}$. So those are removed from $P(B)$.
So $P(B) - P(A) =  ${ {8},{27}, {8,27},{1,8},{1,27}, {1,8,27}}$
A: Let $A$, $B$, $U$, and $U^{'}$ be four sets satisfying
$\quad A \cup B \subset U$
and 
$\quad A \cup B \subset U^{'}$
Then $A \cap B^c = A \cap B^{c^{'}}$, where $^c$ (resp. $^{c^{'}}$  is the complement in $U$ (resp. $U^{'}$).
So if you have only one definition for the set difference operation, and no indication for the universal set, simply set the context yourself and let $U = A \cup B$, defining the universal set to be the minimal set containing both $A$ and $B$.
