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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?

Asaf Karagila
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Victor Jose Arana Rodriguez
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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 Dec 31 '12 at 00:18
  • I agree with that.And I think my anwer is right. Do you agree? – Victor Jose Arana Rodriguez Dec 31 '12 at 00:22
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    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 00:39
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    @AKE: Your edit (and comment) are really strange. What does Russell's paradox has to do with that? – Asaf Karagila Dec 31 '12 at 00:43
  • @AKE Yeah How does this relate to the Russell's paradox? – Victor Jose Arana Rodriguez Dec 31 '12 at 00:45
  • Even if we assumed $S={x:x\not\in x}$ were a set, we would be able to decide this problem; $\wp S$ contains $\varnothing$ and $\varnothing\not\in\varnothing$ (whence $\varnothing\in S$), so $\wp S-S\ne\wp S$. – yearning4pi Dec 31 '12 at 00:48
  • 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 Consider all these withdrawn! – Assad Ebrahim Dec 31 '12 at 00:50

3 Answers3

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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 $\mathbb N$
Community
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Asaf Karagila
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  • What application does it have to define numbers as sets? – Victor Jose Arana Rodriguez Dec 31 '12 at 00:41
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    For example it allows developing all mathematics within a universe of sets. You don't need to have different kind of objects (sets, natural numbers, rational numbers, real numbers, what have you there in your mathematical universe). You only need the empty set and the axioms of set theory. – Asaf Karagila Dec 31 '12 at 00:42
  • @amWhy: Thanks! I find that MSE has a lot of relevant information, often better-written (in my opinion) than Wikipedia. There are probably a couple more links to add here, but these two should suffice for a starting point. – Asaf Karagila Dec 31 '12 at 00:47
  • @AsafKaragila Thank you for the links =) – Victor Jose Arana Rodriguez Dec 31 '12 at 00:50
  • Asaf: I'm just embarrassed that I jumped in too quick without THINKING! Maybe it's time for me to get some sleep! – amWhy Dec 31 '12 at 00:50
  • @amWhy: Maybe it is. I know it's time for me to go to bed, I have to teach in a few hours. – Asaf Karagila Dec 31 '12 at 00:56
  • @Asaf: I'll second amWhy on my response as well. (And apologies to amWhy as well.) – Assad Ebrahim Dec 31 '12 at 00:57
  • @AKE - no need for apologies. All is good. We all make mistakes. I do not know of any user here who has not erred and/or needed to delete a post! And some of the most critical users here don't have the guts to post many answers: they're like vultures hovering to feed off of others, adding little of value to the site. – amWhy Dec 31 '12 at 01:04
  • @amWhy: cheers for that. Thx. – Assad Ebrahim Dec 31 '12 at 01:08
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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 $\wp S-S\ne\wp S$.

yearning4pi
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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 ${\cal P}(S)$.

nonpop
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