If $S$ is a set, let $S^c = \{ x | x \notin S \}$ be the complement of $S$.

In ZFC, if $S$ is any set, then $S^c$ doesn't exist because $S \cup S^c$ would be the universal set, which does not exist in ZFC because of Cantor's paradox and Russell's paradox.

I've always found this statement unusual. Intuitively, I've thought of this result as meaning that all sets in some sense originate at the empty set and "grow upward" to include more and more elements, as opposed to starting with the universal set and "growing downward" by removing elements.

Is there a standard, alternative set of axioms for set theory in which set complements do actually exist for all sets? Or at least "many" sets (for some definition of "many?"

  • $\begingroup$ Why would you want the complements of all sets to exist? It may really be more trouble than it is worth. In classical analysis, for example, you only ever need the complement of a set with respect to some superset of that set. The complement of the set of even numbers, for example, is usually assumed to be its complement with respect to the set of all natural numbers, i.e. the odd numbers. Mathematicians are not usually interested in a sets consisting everything in the universe but the even numbers. $\endgroup$ – Dan Christensen Dec 26 '15 at 4:55
  • $\begingroup$ @DanChristensen Just pure curiosity. I understand that the limited complement makes a lot of sense and I use it all the time. $\endgroup$ – templatetypedef Dec 26 '15 at 15:08
  • $\begingroup$ Also be curious as to why some solutions work so well and while others do not. $\endgroup$ – Dan Christensen Dec 28 '15 at 16:10

In New Foundations (NF) and its variant NFU, all complements exist. NFU is presented in a more elementary way in Randall Holmes’ book Elementary Set Theory with a Universal Set, with an explicit “Axiom of Complements” (p. 19).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.