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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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    Sorry -- I'd interpreted Victor's "yes" to in a comment as referring to the question "is this argument anything to do with Russell's Paradox" -- re-reading it (and the confusion above), that's probably not the case. Consider all these withdrawn!2012-12-31

3 Answers 3

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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.


Some reading material about numbers as sets:

  1. In set theory, how are real numbers represented as sets?
  2. Set theoretic construction of the natural numbers
  3. The history of set-theoretic definitions of $\mathbb N$
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    @amWhy: cheers for that. Thx.2012-12-31
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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 $\wp S-S\ne\wp S$.