5
$\begingroup$

A 2-tuple is $(a, b) = \{\{a\}, \{a, b\}\}$
What about a 1 tuple? Is it $(a) = \{\{a\}\}$?
Then if $a = b$: $(a, b) = (a, a) = \{\{a\}, \{a, a\}\} = \{\{a\}, \{a\}\} = \{\{a\}\}$
So $(a, a) = (a)$?

$\endgroup$
1
  • 3
    $\begingroup$ In the future, please typeset your question with mathjax and add more details about the context behind your question and what you've tried. $\endgroup$ Aug 28, 2021 at 3:15

2 Answers 2

14
$\begingroup$

The definition you are using for an ordered pair is called a Kuratowski pair.

$$ (a, b)_K = \{\{a\}, \{a, b\}\} $$

Unfortunately, it does not immediately generalize beyond 2 elements.

For example, $\{\{a\}, \{a, b\}, \{a, b, c\}\}$ is not a viable definition of an ordered triple because it cannot distinguish $(a, a, b)$ and $(a, b, b)$. This doesn't directly address the singleton case, but it does show that the Kuratowski pair is not a construction for encoding tuples as sets.

One thing you can do if you want to represent tuples of arbitrary length is to represent them as partial functions* with domain $\mathbb{N}$, for example.

$$ (a,) = \{(1, a)_K\} \\ (a, b) = \{(1, a)_K, (2, b)_K \} \\ (a, b, c) = \{(1, a)_K, (2, b)_K, (3, c)_K\} \\ \textit{and so on} $$


Note that in this definition, we're using the Kuratowski pair construction to build our definition of a 2-tuple $(a, b) = \{(1, a)_K, (2, b)_K\}$ rather than directly using the pair as our definition of a 2-tuple.

*Also note that, if we want to define an $n$-ary "function" as a set of $n+1$-tuples where every element of the domain is associated with a unique element of the codomain, then tuples themselves are, strictly speaking, not functions. However, the same intuition from the encoding of a function carries over to the encoding of tuples.

$\endgroup$
5
  • 7
    $\begingroup$ @M.Logic: It's not an error to say Kuratowski's pairing does not generalize to higher lengths. It doesn't. We can apply it itetatively for finite sequences, true. But that is not a generalization. $\endgroup$
    – Asaf Karagila
    Aug 28, 2021 at 7:25
  • $\begingroup$ @Asaf Karagila Maybe we agree. What I mean is that the definition for $n$ tuples is also based on the definition for ordered pairs in the first definition for $n$ tuples in my post. $\endgroup$
    – M. Logic
    Aug 28, 2021 at 7:53
  • 3
    $\begingroup$ I'm careful to distinguish a "Kuratowski pair" $(a, b)_K$ from a "2-tuple" $(a, b)$, but that is a good point. If you want to be able to encode non-unary functions, then you need to encode a function as a set of tuples and thus tuples themselves would not be functions. (I suppose you could use a Kuratoski pair for unary functions only, but that seems a bit odd.) $\endgroup$ Aug 28, 2021 at 8:02
  • 2
    $\begingroup$ Maybe you did it for the benefit of the OP, but starting to count with $1$ instead of $0$… I agree with everything else. $\endgroup$
    – Carsten S
    Aug 28, 2021 at 11:31
  • $\begingroup$ I started with $1$ because I call the projections out of a tuple $\pi_1, \pi_2, \cdots$, not $\pi_0, \pi_1, \cdots$. Is it more usual to start with $0$ (and possibly let the projections be "mismatched")? $\endgroup$ Aug 28, 2021 at 15:06
5
$\begingroup$

In set theory, we usually (see page 18 in Hrbacek, K. and Jech, T., 1999 and page 7 in Jech, T., 2000) define $n$ tuples as follows.

Definition 1. Suppose $n$ is a natural number and $x_0,\cdots,x_{n-1}$ are sets. Set

\begin{align*} (\,)&:=\varnothing,\\ (x_0)&:=x_0,\\ (x_0,x_1)&:=\{\{x_0\},\{x_0,x_1\}\},\\ \vdots\quad\qquad&\qquad\qquad~\vdots\\ (x_0,\cdots,x_{n-2},x_{n-1})&:=((x_0,\cdots,x_{n-2}),x_{n-1}). \end{align*}

Remark 2. As you see, $0$ tuples and $1$ tuples has nothing to do with Kuratowski's definition for ordered pairs$^*$, while $n$ tuples with $n\geq 2$ are based on Kuratowski's definition for ordered pairs. The reason that we set $0$ tuples to be the empty set is clear, and the reason that we set $1$ tuples to be the element laying in itself is to coincide with definitions for unary relations (any subset of $X$ is a unary relation on $X$) and with definitions for unary Cartesian products ($X^1=X$ which selects elements of itself as its elements).

Furthermore, keeping the definitions for $0$ tuples, $1$ tuples and $2$ tuples unchanged, we can also define $n$ tuples with $n\geq 3$ to be functions with domain $n$.

Definition 3. Suppose $n$ is a natural number no less than $3$ and $x_0,\cdots,x_{n-1}$ are sets. Set

$$(x_0,\cdots,x_{n-1}):=\{(0,x_0),\cdots,(n-1,x_{n-1})\}.$$

Remark 4. Note that $2$ tuples (ordered pairs$^\dagger$) can't be defined as functions with domain $2$, since the definitions for functions depends on relations while the definitions for relations depends on $2$ tuples, and so the definition for $2$ tuples should only be Kuratowski's or similar (for example, Hausdorff's or Wiener's) way.

Remark 5. Both of the two definitions for $n$ tuples with $n\geq 3$ have their own advantages: (1) Definition 1 has real order while Definition 3 has imaginary order since the elements of functions have no order and the imaginary order follows from the order of ordinals (natural numbers); (2) Definition 3 is clearly simpler than the first one. In fact, since we also define sequences as functions with ordinals as domains, if we adopt Definition 1, then tuples can only be regarded as finite sequences while in fact not, and if we adopt Definition 3 then $n$ tuples are indeed $n$ sequences (finite sequences).

Anyway, $(a,a)=\{\{a\}\}\neq a=(a)$. Note that by Axiom of Foundation/Regularity, we can by induction show that: if $n$ is a non-zero natural number, then $a\neq \underbrace{\{\cdots\{}_{n\text{ many}}a\underbrace{\}\cdots\}}_{n\text{ many}}$.


$*$ By the way, Kuratowski's definition for ordered pairs is great although it seems to be very simple at first glance since it transforms the order of elements laying in the ordered pairs into the difference of sets (not difference operation of sets) of elements laying in the ordered pairs, and based on it we can define other basic mathematical concepts such as relations, functions and so on.

$\dagger$ Considering of no mathematical definition, there are many equivalent names to $2$ tuples: couple, double, ordered pair, two-ple, twin, dual, duad, dyad, twosome. So it's better to define $2$ tuples and ordered pairs to be the same object, while @Gregory Nisbet 's new definition for $(a,b)$ (i.e., 2 tuples) and definition for $(a,b)_K$ (i.e., Kuratowski's definition for ordered pairs) are different objects. Without this, it's also ok to define $2$ tuples in @Gregory Nisbet 's new way. By the way, in fact @Gregory Nisbet 's new way is the same as Definition 3 for $n$ tuples with $n\geq 3$ as above.

$\endgroup$
15
  • 3
    $\begingroup$ We do??? I always thought that we define ordered pairs, and then tuples are just functions with the appropriate domain. $\endgroup$
    – Asaf Karagila
    Aug 28, 2021 at 7:09
  • 1
    $\begingroup$ Yes, but what you're saying is that $((((x)))) = (((x))) = ((x)) = (x) = x$, and worse, $(x, y) =((x, y))$. So, is it a pair or a tuple? $\endgroup$
    – Asaf Karagila
    Aug 28, 2021 at 7:23
  • 2
    $\begingroup$ I find the statement that 2-tuples can't be defined as functions misleading: one defines the ordered pair $(a,b)$ using Kuratowski's definition, and then define an $n$-tuple as a set of ordered pairs containing exactly one pair $(k,x)$ for every $k\in n$. This implies that 2-tuples and ordered pairs are different, but I don't see a problem there. It feels a lot more natural than the idea that 2-tuples somehow can't be defined as functions, while the rest can. $\endgroup$
    – Vsotvep
    Aug 28, 2021 at 7:45
  • 2
    $\begingroup$ @M.Logic It would however be more satisfying to have definitions that "work" also in $\mathsf{ZF}-$Foundation, say. $\endgroup$ Aug 28, 2021 at 13:57
  • 2
    $\begingroup$ @AsafKaragila So, is it a pair or a tuple? – this definition of n-tuples performs ‘type erasure’: in other words, given this definition of tuples, when you get handed a set, you have to know in advance which kind of tuple it is supposed to be in order to meaningfully operate on it (or obtain this information from some other object, a form of ‘RTTI’ if you will). $\endgroup$ Aug 28, 2021 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.