# 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?

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Well, $\wp S$ is a set of sets, while $S$ is a set of "elements". We have $S\in \wp S$; whilst $S\subseteq S$. – Pedro Tamaroff Dec 31 '12 at 0:18
I agree with that.And I think my anwer is right. Do you agree? – Victor Jose Arana Rodriguez Dec 31 '12 at 0:22
That depends on what 0, 1, ... Are in your eyes. If you are doing set theory, the natural numbers may have been introduced as sets derived from the empty set. Depending on the details, the intersection of $S$ and $\wp S$ may not be empty. – mkl Dec 31 '12 at 0:39
@AKE: Your edit (and comment) are really strange. What does Russell's paradox has to do with that? – Asaf Karagila Dec 31 '12 at 0:43
@AKE Yeah How does this relate to the Russell's paradox? – Victor Jose Arana Rodriguez Dec 31 '12 at 0:45

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 ${\cal P}(S)$.

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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.

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 $\wp S-S\ne\wp S$.