Does a complement contain itself?

Question

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?

Answer 1

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.

Answer 2

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