1.2Disjoint Unions and Cartesian Products

Updated 11 May 2026Chapter (PDF)

A word about disjoint unions and Cartesian products is in order.

[1.2.1]Definition(Cartesian Product)#

The Cartesian product of two sets $A, B$ is defined by

A \times B = \{(a, b) : a \in A, b \in B\}.

[1.2.2]Remark(Cartesian Product is not Canonical)#

For two sets $V_1, V_2$ , their Cartesian product is uniquely determined—it is just a set of all the pairs of elements in $V_1$ and $V_2$ . But consider the Cartesian product of more than two sets, say $V_1 \times V_2 \times V_3$ . We can interpret it in two ways:

V_1 \times (V_2 \times V_3) = \{(v_1, (v_2, v_3)): v_k \in V_k\},

(V_1 \times V_2) \times V_3 = \{((v_1, v_2), v_3): v_k \in V_k\}.

Effectively, the Cartesian product is not associative. Thus $V_1 \times \dots \times V_n$ is ill-defined as a set since a choice needs to be made.

[1.2.3]Definition(Disjoint Union of Sets)#

Suppose $S, T$ are two sets. Let $S', T'$ be such that $S \cong S'$ , $T \cong T'$ , and $S' \cap T' = \emptyset$ . Then the disjoint union of $S, T$ is $S \amalg T = S' \cup T'$ .

[1.2.4]Remark(Disjoint Union is not Canonical)#

Of course, the disjoint set union is not canonical—the isomorphic sets used in the expression are chosen arbitrarily. One can let $S' = S \times \{0\}$ and $T' = T \times \{1\}$ for a convenient choice, but, of course, this is a choice.

We finish by noting that the Cartesian product (of three or more sets) and disjoint union of sets are not canonically chosen and are technically undefined as unique sets. However, any choices made in their construction lead to isomorphic candidates, meaning that the results of these operations are well-defined up to isomorphism. The main feature of constructions like products and disjoint unions is not really “what elements they contain” but rather “their relationship with all other sets”.