Power set difference on the same set. I've been arguing about the following expression:
Given the following set $ S := \{1,2,3,4,5\}$ evaluate the expression:
$$ \wp S - S = $$
I think that the result is 
$$\wp S - S = \wp S $$
Because $\wp S$ and $S$ don't have elements in common. Am I right?
 A: That really depends on what are the elements of $S$. It is common in introductory courses to assume that numbers are not sets, in which case it is clear that $\wp(S)\setminus S=\wp(S)$, because every element in $\wp(S)$ is a set, whereas every element of $S$ is not a set.
However it is also a common practice to define numbers by sets: $$0=\varnothing; 1=\{0\}; 2=\{0,1\}; 3=\{0,1,2\}; 4=\{0,1,2,3\}; 5=\{0,1,2,3,4\}; 6=\{0,1,2,3,4,5\}.$$
In this case $S=6\setminus\{0\}$. One can calculate and see that in such case $\wp(S)$ contains all the elements of $S$, and then one has to sit down and write in slightly more details which sets remain in $\wp(S)$ after the difference. 
However if $x\in S$ then $0\in x$, but $0\notin S$, and therefore $x\nsubseteq S$. It follows that $S\cap\wp(S)=\varnothing$, again. This presentation is the von Neumann ordinals. One can use Zermelo's representation, which is as follows: $$0=\varnothing; 1=\{\varnothing\}; 2=\{\{\varnothing\}\}; 3=\{\{\{\varnothing\}\}\}; 4=\{\{\{\{\varnothing\}\}\}\}; 5=\{\{\{\{\{\varnothing\}\}\}\}\}.$$
In this presentation, $S$ does have common elements as $\wp(S)$, e.g. $5=\{4\}$, and therefore $5\in\wp(S)$. So in this case $2,3,4,5\in S\cap\wp(S)$, and therefore $\wp(S)\setminus S$ has $2^5-4$ elements which one can calculate by hand.
Zermelo's interpretation can still be found in some places today, although it's less common because we cannot extend it in a natural way to transfinite ordinals, whereas von Neumann's interpretation carries on just fine.

Some reading material about numbers as sets:


*

*In set theory, how are real numbers represented as sets?

*Set theoretic construction of the natural numbers

*The history of set-theoretic definitions of $\mathbb N$
A: The powerset of $S$ is the set of all $T\subseteq S$. If $\wp S-S$ were to be different from $\wp S$, then at least one element of $S$ is a subset of $S$. Depending on how you define ordinals, this may or may not be true.
E.g. say $2=\{1\}$. Then $2\subseteq S$, so $\wp S-S\ne\wp S$.
A: Let's try it with a smaller set. Let $S=\{1,2\}$. Then ${\cal P}(S)=\{\emptyset,\{1\},\{2\},S\}$. Now ${\cal P}(S)\setminus S$ contains exactly those elements which are in ${\cal P}(S)$ but not in $S$. Thus $${\cal P}(S)\setminus S=\{\emptyset,\{1\},\{2\},S\}\setminus\{1,2\}={\cal P}(S)$$
because neither $1$ or $2$ is an element of ${\cal P}(S)$.
