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I think it is safe to assume many sets do not contain their complements. {1, 2} for example. Now, by the definition of complement, that would mean the complement would have to contain itself. This violates the axiom of regularity. What is the resolution to this paradox?

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closed as unclear what you're asking by Jonas Meyer, Claude Leibovici, Joel Reyes Noche, drhab, Chappers Jun 10 at 10:08

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What do you mean by the complement of $\{1,2\}$? –  Brian M. Scott Jan 5 '13 at 18:05
When you talk of "complement of a set", you must make clear within what universal set you're working. –  DonAntonio Jan 5 '13 at 18:07
Note that a set has empty intersection with its complement. There is no paradox. I can't see exactly what you are trying to do here, but it looks to me as you may be mixing a set up with its power set. –  Geoff Robinson Jan 5 '13 at 18:10

2 Answers 2

Assuming by "contain" you mean $\ni$, your problem is that you seem to be assuming that if anything is not an element of a set $A$, then it must be an element of its complement $A^c$. But such a set $A^c$ would have to contain all sorts of things, including arbitrary other sets, equilateral triangles, Jimmy Page, invisible pink unicorns, and so on.

Joking aside, under this interpretation, the complement of the empty set would be a universal set that contains everything, and such a set cannot exist in the usual formulations of set theory. So the complement of a set is only well-defined if you first specify a universal set $U$ of objects you are considering, then define the complement of $A$ as the set $A^c = U \setminus A$, the set of elements of $U$ that are not in $A$.

For example, if you're working in $\mathbb Z$, the set of integers, then the complement of $A = \{1,2\}$ is $A^c = \mathbb Z \setminus A = \{\ldots, -2, -1, 0, 3, 4, 5, \ldots\}$. Does this set contain itself? No.

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Well, any set whatsoever always contains itself, by definition of set contention, thus in particular the complement of a set contains itself...nothing surprising.

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I think the OP means $\ni$, not $\supset$. –  mrf Jan 5 '13 at 23:35
"Contention" meaning what, $\supseteq$ ?? –  GEdgar Jan 6 '13 at 1:22
Yes @GEdgar, and though this may be a language barrier, I think the meaning is clear from the context in my answer: for any set $\,X\,$ it is true that $\,X\subset X\,$ –  DonAntonio Jan 6 '13 at 2:45

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