ZFC Axioms

Historical Background

It would be no exaggeration to say that set theory, as the foundation of mathematics, originated with Cantor in the 19th century. Cantor’s naive set theory rests on the following principle:

For any property \(P\), there exists a set \(Y=\{x\mid P(x)\}\) consisting of all elements satisfying \(P\).

However, mathematicians soon discovered that this approach leads to various contradictions.

Example 1 (Russell’s Paradox) Consider the collection \(\mathcal{S}\) of all \(x\) such that \(x\not\in x\). Then \(\mathcal{S}\) is either an element of itself or not.

• Suppose \(\mathcal{S}\) is an element of itself. Then \(\mathcal{S}\) must satisfy the defining property (\(x\not\in x\)), which implies \(\mathcal{S}\not\in\mathcal{S}\). This contradicts the assumption that \(\mathcal{S}\in\mathcal{S}\), so \(\mathcal{S}\) cannot be an element of itself.
• It follows that \(\mathcal{S}\not\in\mathcal{S}\). Yet this too leads to a contradiction: since \(\mathcal{S}\) is the collection of all elements satisfying \(x\not\in x\), and \(\mathcal{S}\) satisfies this property, it must belong to \(\mathcal{S}\).

Thus, whether \(\mathcal{S}\) is an element of itself or not, a contradiction arises. This is called Russell’s paradox.

To avoid such paradoxes, sets must be defined rigorously rather than merely as “collections of elements.” The framework that accomplishes this is axiomatic set theory.

The posts in the Set Theory category are not intended as an introduction to axiomatic set theory per se; however, as a starting point, it seems fitting to provide a brief overview of the ZFC axiom system, which is the most widely adopted framework.

ZFC Axioms

If no sets existed at all, then any statement about sets would be vacuously true. This follows from the fact that the logical formula \(P\implies Q\) is always true when \(P\) is false, regardless of the truth value of \(Q\). Our first axiom therefore asserts that at least one set exists.

The Axiom of Existence. There exists a set with no elements.

To refer to this “set with no elements” as the empty set, we must first establish its uniqueness. The following axiom provides the justification.

The Axiom of Extensionality. For two sets \(A,B\), if every element of \(A\) is an element of \(B\) and every element of \(B\) is an element of \(A\), then the two sets are equal.

Proposition 2 There exists at most one set with no elements.

Proof

Let \(A\) and \(B\) be sets with no elements. Then both statements

\[((x\in A)\implies (x\in B)),\qquad ((x\in B)\implies (x\in A))\]

are true. Therefore, by the axiom of extensionality, \(A=B\).

Since the axiom of existence tells us that at least one such set exists, both the existence and uniqueness of a “set with no elements” are guaranteed. We now call this set the empty set and denote it by \(\emptyset\).

The following axiom is worth remembering as it prevents Example 1.

The Axiom Schema of Comprehension. Given any set \(A\) and any proposition \(P\), there exists a set \(B\) such that \(x\in B\) is equivalent to (\(x\in A\) and \(P(x)\)).

Formally, the above axiom asserts that a certain property holds for all propositions \(P\). Since this cannot be expressed in first-order logic as a single axiom, we regard it as a collection of axioms rather than a single one—hence the term axiom schema of comprehension.

The set \(B\) defined by the comprehension schema is unique. If \(B'\) is another set satisfying this condition, then

\[x\in B'\iff ((x\in A)\wedge P(x))\iff x\in B\]

and thus \(B=B'\). We denote this set by \(\{x\in A\mid P(x)\}\).

Example 3 The source of contradiction in naive set theory was the following assumption:

Let \(P\) be a proposition about \(x\). Then there exists a set \(B\) such that \(x\in B\) is equivalent to \(P(x)\).

Under the comprehension schema introduced above, unlike in Example 1, we cannot directly define \(\mathcal{S}=\{x\mid x\not\in x\}\); we can only define

\[B=\{x\in A\mid x\not\in x\}\]

for an already existing set \(A\). Unlike the set \(\mathcal{S}\), this set yields no contradiction.

• If \(B\in B\), then by definition \(B\not\in B\) and \(B\in A\), which is a contradiction.
• If \(B\not\in B\), then \(B\not\in A\) or \(B\in B\).

In the second case, if \(B\in B\) then we obtain a contradiction, so necessarily \(B\not\in A\). This blocks the contradiction that arises in Russell’s paradox.

Example 4 The preceding example demonstrates that for any set \(A\), there always exists a set \(B\) that does not belong to \(A\). In particular, the “set of all sets”—that is, a universal set—does not exist.

By choosing the proposition \(P\) appropriately, we can construct various familiar sets using the comprehension schema. Their uniqueness follows immediately from extensionality.

Example 5 For any sets \(A\) and \(B\), let the property \(P\) concerning \(x\) be given by \(x\in B\). Then the set

\[\{x\in A\mid P(x)\}\]

is precisely the collection of elements belonging to both \(A\) and \(B\). We call this set the intersection of \(A\) and \(B\), denoted \(A\cap B\).

Example 6 Now, given two sets \(A\) and \(B\), let the property \(Q\) concerning \(x\) be given by \(x\not\in B\). Then the set

\[\{x\in A\mid Q(x)\}\]

consists of all elements belonging to \(A\) but not to \(B\). We call this the difference of \(B\) from \(A\), denoted \(A\setminus B\).

The set \(A\setminus B\) is also referred to as the complement of \(B\) relative to \(A\).¹

Without guaranteeing the existence of sets beyond the empty set, the above two examples would be of little use. The following axioms provide methods for constructing non-empty sets.

The Axiom of Pair. For any sets \(A\) and \(B\), there exists a set \(C\) such that \(x\in C\) if and only if (\(x=A\) or \(x=B\)).

Again, this set is unique by extensionality and is denoted \(\{A,B\}\). Now taking \(A=B=\emptyset\), from

\[x\in \{\emptyset\}\iff x=\emptyset\iff (x=\emptyset)\wedge(x=\emptyset)\iff x\in \{\emptyset,\emptyset\}\]

we obtain \(\{\emptyset, \emptyset\}=\{\emptyset\}\). Also, since \(\emptyset\not\in \emptyset\), we have \(\emptyset\neq\{\emptyset\}\).

The Axiom of Union. For any set \(\mathcal{S}\), there exists a set \(U\) such that \(x\in U\) if and only if (\(x\in A\) for some \(A\in\mathcal{S}\)).

For example, if \(\mathcal{S}=\{A,B\}\), then \(U\) is the set of elements satisfying (\(x\in A\) or \(x\in B\)), i.e., \(A\cup B\). This is sometimes written as \(\bigcup\mathcal{S}\).

The Axiom of Power Set. For any set \(S\), there exists a set \(\mathcal{P}\) such that \(X\in \mathcal{P}\) if and only if \(X\subseteq S\).

This set is called the power set of \(S\) and is denoted \(\mathcal{P}(S)\).

The ZFC axiom system includes several additional axioms beyond those presented here, which will be introduced as needed.

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References

[HJJ] K. Hrbacek, T.J. Jeck, and T. Jech. Introduction to Set Theory. Lecture Notes in Pure and Applied Mathematics. M. Dekker, 1978.
[Bou] N. Bourbaki. Elements of the History of Mathematics. Springer, 2013
Wikipedia, Naive set theory, Set-theoretic definition of natural numbers.

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1. In axiomatic set theory, the “complement of \(B\)” does not exist as such, since its definition would require a universal set. ↩