Tuple

Tuples as functions

The ${\displaystyle 0}$-tuple may be identified as the empty function. For ${\displaystyle n\geq 1,}$ the ${\displaystyle n}$-tuple ${\displaystyle \left(a_{1},\ldots ,a_{n}\right)}$ may be identified with the (surjective) function

${\displaystyle F~:~\left\{1,\ldots ,n\right\}~\to ~\left\{a_{1},\ldots ,a_{n}\right\}}$

with domain

${\displaystyle \operatorname {domain} F=\left\{1,\ldots ,n\right\}=\left\{i\in \mathbb {N} :1\leq i\leq n\right\}}$

and with codomain

${\displaystyle \operatorname {codomain} F=\left\{a_{1},\ldots ,a_{n}\right\},}$

that is defined at ${\displaystyle i\in \operatorname {domain} F=\left\{1,\ldots ,n\right\}}$ by

${\displaystyle F(i):=a_{i}.}$

That is, ${\displaystyle F}$ is the function defined by

${\displaystyle {\begin{alignedat}{3}1\;&\mapsto &&\;a_{1}\\\;&\;\;\vdots &&\;\\n\;&\mapsto &&\;a_{n}\\\end{alignedat}}}$

in which case the equality

${\displaystyle \left(a_{1},a_{2},\dots ,a_{n}\right)=\left(F(1),F(2),\dots ,F(n)\right)}$

necessarily holds.

Tuples as sets of ordered pairs

Functions are commonly identified with their graphs, which is a certain set of ordered pairs. Indeed, many authors use graphs as the definition of a function. Using this definition of "function", the above function ${\displaystyle F}$ can be defined as:

${\displaystyle F~:=~\left\{\left(1,a_{1}\right),\ldots ,\left(n,a_{n}\right)\right\}.}$

Tuples as nested ordered pairs

Another way of modeling tuples in Set Theory is as nested ordered pairs. This approach assumes that the notion of ordered pair has already been defined.

1. The 0-tuple (i.e. the empty tuple) is represented by the empty set ${\displaystyle \emptyset }$.
2. An n-tuple, with n > 0, can be defined as an ordered pair of its first entry and an (n − 1)-tuple (which contains the remaining entries when n > 1):

   ${\displaystyle (a_{1},a_{2},a_{3},\ldots ,a_{n})=(a_{1},(a_{2},a_{3},\ldots ,a_{n}))}$

This definition can be applied recursively to the (n − 1)-tuple:

${\displaystyle (a_{1},a_{2},a_{3},\ldots ,a_{n})=(a_{1},(a_{2},(a_{3},(\ldots ,(a_{n},\emptyset )\ldots ))))}$

Thus, for example:

${\displaystyle {\begin{aligned}(1,2,3)&=(1,(2,(3,\emptyset )))\\(1,2,3,4)&=(1,(2,(3,(4,\emptyset ))))\\\end{aligned}}}$

A variant of this definition starts "peeling off" elements from the other end:

1. The 0-tuple is the empty set ${\displaystyle \emptyset }$.
2. For n > 0:

   ${\displaystyle (a_{1},a_{2},a_{3},\ldots ,a_{n})=((a_{1},a_{2},a_{3},\ldots ,a_{n-1}),a_{n})}$

This definition can be applied recursively:

${\displaystyle (a_{1},a_{2},a_{3},\ldots ,a_{n})=((\ldots (((\emptyset ,a_{1}),a_{2}),a_{3}),\ldots ),a_{n})}$

Thus, for example:

${\displaystyle {\begin{aligned}(1,2,3)&=(((\emptyset ,1),2),3)\\(1,2,3,4)&=((((\emptyset ,1),2),3),4)\\\end{aligned}}}$

Tuples as nested sets

Using Kuratowski's representation for an ordered pair, the second definition above can be reformulated in terms of pure set theory:

1. The 0-tuple (i.e. the empty tuple) is represented by the empty set ${\displaystyle \emptyset }$;
2. Let ${\displaystyle x}$ be an n-tuple ${\displaystyle (a_{1},a_{2},\ldots ,a_{n})}$, and let ${\displaystyle x\rightarrow b\equiv (a_{1},a_{2},\ldots ,a_{n},b)}$. Then, ${\displaystyle x\rightarrow b\equiv \{\{x\},\{x,b\}\}}$. (The right arrow, ${\displaystyle \rightarrow }$, could be read as "adjoined with".)

In this formulation:

${\displaystyle {\begin{array}{lclcl}()&&&=&\emptyset \\&&&&\\(1)&=&()\rightarrow 1&=&\{\{()\},\{(),1\}\}\\&&&=&\{\{\emptyset \},\{\emptyset ,1\}\}\\&&&&\\(1,2)&=&(1)\rightarrow 2&=&\{\{(1)\},\{(1),2\}\}\\&&&=&\{\{\{\{\emptyset \},\{\emptyset ,1\}\}\},\\&&&&\{\{\{\emptyset \},\{\emptyset ,1\}\},2\}\}\\&&&&\\(1,2,3)&=&(1,2)\rightarrow 3&=&\{\{(1,2)\},\{(1,2),3\}\}\\&&&=&\{\{\{\{\{\{\emptyset \},\{\emptyset ,1\}\}\},\\&&&&\{\{\{\emptyset \},\{\emptyset ,1\}\},2\}\}\},\\&&&&\{\{\{\{\{\emptyset \},\{\emptyset ,1\}\}\},\\&&&&\{\{\{\emptyset \},\{\emptyset ,1\}\},2\}\},3\}\}\\\end{array}}}$