The ordered triple $(a,b,c):=((a,b),c)$ and the iterated products $A\times B\times C:=(A\times B)\times C$

For sets $a$, $b$, $c$ the ordered triple is

$(a,b,c):=((a,b),c),$

an ordered pair (The Kuratowski ordered pair $(a,b):=\{\{a\},\{a,b\}\}$) whose first coordinate is itself an ordered pair, and for sets $A$, $B$, $C$ the iterated product is

$A\times B\times C:=(A\times B)\times C,$

a Cartesian product (The Cartesian product $A\times B:=\{z\in \mathcal{P}(\mathcal{P}(A\cup B)):\mathrm{\exists }a\in A\mathrm{\exists }b\in Bz=(a,b)\}$) of the same shape. Both conventions associate to the left, and the elements of $A\times B\times C$ are exactly the triples $(a,b,c)$ with $a\in A$, $b\in B$ and $c\in C$.

Applying $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$ twice gives the characterising property of triples: $(a,b,c)=(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$ holds if and only if $a=a^{\mathrm{\prime }}$, $b=b^{\mathrm{\prime }}$ and $c=c^{\mathrm{\prime }}$.

• The bracketing convention is not a formality. $(A\times B)\times C$ and $A\times (B\times C)$ are in general different sets, so a convention has to be fixed and adhered to; the left-associated one is fixed here.

• No $n$-tuples. A general $n$-tuple is a function on a natural number, and the natural numbers are not available at this point in the reading order, so only triples and finitely iterated binary products are introduced here. The general construction is the product of an indexed family, once functions and index sets are available.