# Does universal set exist?

Take any set $A$. Is it true that the complement of $A$ is non empty?

I am unable to find which axiom in set theory leads to answer affirmative of this question.

• Other may explain this better than I can. The Russell Paradox means that "the universal set" cannot exist as naively conceived. But there is a second question about whether there is a set which has $A$ as a subset and also contains an element not in $A$ - one way of solving the Russell Paradox problem is to make sure that everything you do is contained within some well-defined large set - which may be a different set for different problems or constructions. – Mark Bennet Jun 5 '13 at 9:59

## 4 Answers

The complement is only formed with respect to another set. This means the things (also sets, usually) in the compliment should be elements of other sets.

Let's start from below: If $A$ is some set, you can try to form the set which contains $A$'s elements and some more beyond it, e.g. $X=A\cup \{a\}=\{x|x\in A \lor x=a\}$. And the compliment of $A$ with respect to this $X$ is nonempty if $a$ isn't in $A$ already.

The details of which sets are provable to exist, and that are not already in $A$, will depend on the set theory you work with. Specifically it will depend on which existence axioms enable you to construct new sets, possibly via using a given set such as $A$. Check for example the kind of axioms like Comprehension via..., which are part of theories involving sets. Here I said "axioms which enable you to construct", but I should be more cautious and say "axioms which let you show existence of". Here is a list with some set theories. This set theory doesn't even contain the power set of all its sets. If the theory lets you put $A$ in another set with other sets (with different elements) next to it, you can form the union to get a bigger set so that the compliment of $A$ w.r.t it will be nonempty.

I don't know of any reference of someone trying to write down a set theory which e.g. only contains $\emptyset$. In this case there is no non-empty compliment. Maybe you can take a set theory and drop axioms until you get a stronger (i.e. not completely trivial) finite theory with a biggest set, such that you can't construct a bigger set. Maybe you might achieve this by listing unpractical axioms granting the existence of some sets but not possibility to create new and different ones.

If you merely write down the axioms for your set theory, it'll usually not be immediately obvious to which mathematical objeect (the sets) the theory gives you access. That is one theory will usually contain objects which are not in another theory, i.e. have no clear counterpart in another theory. Some notable standard big sets are found here: Constructible universe.

For simplicity, let me assume that you intend to ask your question in the context of the $\mathsf{ZF(C)}$ axioms.

There are sets that are not elements of $A$ (otherwise we would run into Russell's paradox).

If we use the axioms of Pairing and Regularity, then we obtain that $A \notin A$:

By Pairing applied to $A$ with itself, obtain $\{A\}$. Now by Regularity $\{A\}$ contains an element disjoint from itself; therefore, $A \notin A$.

However, it is important to understand that speaking of "the complement of $A$" is not as well-behaved as one would like to think.

For, Regularity proves that $\{x, A\} \notin A$ for all sets $x$; thus the tentative complement of $A$ is "too big" to possibly be a set, and we speak of a proper class.

In mathematical practice, it is customary to deal with the complement of $A$ only when $A$ is given as a subset of some larger set $U$. In this sense, we generally only discuss the relative complement $A^c_U = U \setminus A$; it has the pleasant property that it is always a set, and for most purposes, this approach suffices.

In this context, $U$ may be referred to as "the universe of discourse", a set that contains everything we like to talk about in our context.

For each set $A$ let $$r(A) = \{ x \in A : x \notin x \}.$$ By Separation $r(A)$ exists. It is easy to see that $r(A) \notin A$. Suppose not, then $$r(A) \in r(A) \Leftrightarrow r(A) \notin r(A),$$ which is a contradiction. Therefore for any set A, there exists a set $r(A) \in A^c$. It should be noted of course that $A^c$ is a proper class, otherwise $V$ the set of all sets exists and we get a contradiction by Russell's Paradox, i.e. $r(V) \in V = r(V)$, which as we saw is a contradiction.

It depends on your set theory. Whitehead and Russell's theory of classes in Principia Mathematica has a universal set $V$ (actually, several such) whose complement is empty. Other set theories exist in which there are universal sets. You may be interested to read Randall Holmes' monograph on elementary set theory with a universal set, which is an exposition of Quine and Jensen's set theory, called NFU. In NFU, there is a universal set, and its complement is empty.

Other set theories do not have a universal set. The most commonly-used system for set theory, ZF, does not have a universal set: from its Axiom of Foundation, one can prove that for any $x$, there exists $y$ with $y\notin x$. (A simple example is $x$ itself.) But in ZF there are no absolute complements. You can construct the set of elements of $X$ that are not in $Y$, and this is the "relative complement" $X\setminus Y$. But there is no way to construct the set of everything that is not in $Y$, precisely because there is no "everything" to construct it from.