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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!

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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.

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  • $\begingroup$ I am given with just the two sets A and B. And I needed to work on the complement of A so I assumed I need a universal set. I also need to list down the elements of P(B-A) and P(B)-P(A). $\endgroup$ – frstrtdmthmtcn Oct 30 '17 at 23:55
  • $\begingroup$ I see. If no context is provided, you will have to choose a universal set (and clearly state it before you proceed). Are you sure there's no contextual information that indicates a certain universal set? Sometimes this is assumed from the definition of the sets, e.g if I say "let $X,Y \subseteq \mathbb{R}$" with no additional information, you will assume $\mathbb{R}$ is the universal set here. $\endgroup$ – Guido A. Oct 31 '17 at 0:14
  • $\begingroup$ 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)." $\endgroup$ – frstrtdmthmtcn Oct 31 '17 at 0:27
  • $\begingroup$ @frstrtdmthmtcn: Why do you need a universal set for that? $\endgroup$ – Asaf Karagila Oct 31 '17 at 0:27
  • $\begingroup$ To get the complement of P(A). Is it wrong? $\endgroup$ – frstrtdmthmtcn Oct 31 '17 at 0:28
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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 $B$.

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"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) = ${ {8},{27}, {8,27},{1,8},{1,27}, {1,8,27}}$

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  • $\begingroup$ What if I have sets A and B and the universal set U. If I get the complement of A x B, should I use U x U as basis? $\endgroup$ – frstrtdmthmtcn Oct 31 '17 at 1:43

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