Defining “tuples” of $n \geq 2$ as SETS of ordered pairs (indexing strategy) versus single ordered pairs (Kuratowski strategy)

elementary-set-theory

After reading some posts on the forum (Definition of the ordered triple (a, b, c) according to Kuratowski's Set Theory. and Associativity of cartesian product and nested ordered n-tuples to name a few), I learned that there are two common ways of defining elements of Cartesian Products of the form $X \times X \times X \times …$

The first way involves purely applying the Kuratowski definition (recursively if $n \gt 2$), which will produce a single ordered pair, and the second way involves an indexed set ( $\{0,1,…,n\}$) mapping, which ends up producing a collection (set) of ordered pairs.

Let $X \times X$ and $X \times X \times X$ be my two working examples.

Consider an element $\langle a,b \rangle \in X \times X$

"Purely" Kuratowski Definition – $\langle a, b \rangle := \{\{a\},\{a,b\}\}$

Indexed Set Definition (followed by Kuratowski) – $\langle a,b \rangle := \{\langle 0,a\rangle,\langle1,b\rangle\} = \Big \{\big \{\{0\},\{0,a\}\big\}, \big \{\{1\},\{1,b\}\big\}\Big\}$

Consider an element $\langle a,b,c \rangle \in X \times X \times X$

"Purely" Kuratowski Definition – $\langle a, b, c \rangle := \langle \langle a, b \rangle, c \rangle = \Big \{ \big \{ \big \{ \{a\},\{a,b\}\big \} \big \}, \big \{ \big \{ \{a\}, \{a,b \} \big \}, c \big \} \Big \}$

Indexed Set Definition (followed by …) – $\langle a, b, c \rangle :=\{\langle 0,a\rangle,\langle1,b\rangle\,\langle 2,c\rangle\} = \Big \{\big \{\{0\},\{0,a\}\big\}, \big \{\{1\},\{1,b\}\big\},\big \{\{2\},\{2,c\}\big\}\Big\}$

Now, my question is as follows:

For a given Cartesian Product "dimension" ($2$ in the case of $X \times X$ and $3$ in the case of $X \times X \times X$), are these definitions "bijective" (maybe isomorphic is the appropriate term here)?

I am guessing the answer is yes, but I am not really seeing what that might look like. I've always thought of bijective functions of finite sets as requiring each set to have the same number of elements. In the $2$D example, this is certainly true…but in the $3$D example, this is not true. So I am uncertain of what to think.

Any insight is appreciated! Thank you.

Best Answer

In each case one can uniquely recover the components and their order from the $n$-tuple. Suppose, for example, that $t\in A\times B\times C$ is a Kuratowski ordered triple. Then there are a unique $p\in A\times B$ and $c\in C$ such that $t=\langle p,c\rangle$, and there are unique $a\in A$ and $b\in B$ such that $p=\langle a,b\rangle$. Thus, we can read off the first, second, and third components of $t$ as $a,b$, and $c$, respectively: $t=\langle a,b,c\rangle$.

Mind you, actually describing how to recover $a$ and $b$ from $p$, for instance, is a bit of a pain, but it can be done. First, $a$ is the unique $x\in\bigcup p$ such that $\forall y\in p\,(x\in y)$. If $a$ is the only element of $\bigcup p$, then $b=a$; otherwise, $b$ is the unique element of $\left(\bigcup p\right)\setminus\{a\}$.

From the other definition, of course, we recover $a,b$, and $c$ as $t(0),t(1)$, and $t(2)$, respectively. The bijection between the two versions of $A\times B\times C$ is then defined in terms of the components: if $\langle a,b,c\rangle$ is a Kuratowski triple, the indexed set triple corresponding to it under the bijection is the unique $t$ in the indexed set product such that $t(0)=a$, $t(1)=b$, and $t(2)=c$.

All of this generalizes readily to $n$-fold products for arbitrary $n\in\Bbb Z^+$.

Related Question