2
$\begingroup$

I have been doing some reading on set theory and I have come across "ordered pairs". I understand that sets in general are unordered, and when we want to place them in a particular order we use the 'Kuratowski' definition to say, for example a pair of elements such as $(a,b)=\{(a),(a,b)\}$ which is why $a$ is the 'first entry' and $b$ the 'second entry'.

I also understand that ordered pairs are very important and are the very definition of coordinate systems.

What I want to understand is, say we wanted to order a set of $3,4...n$ elements, how would we do that, and why would we want to do that even at all?

I've read that those are called ordered $n-tuples$, but they seemed to have just divided the set into two elements, bunching up all elements save one into a subset, how would we even interpret something like that? e.g

$(a,b,c,...,z)=\{(a,b,c...,y), (z)\}$

That confuses me a little because you haven't really ordered anything. All you've done was make the last element $z$. The first $25$ elements are still ambiguous.

$\endgroup$
  • 2
    $\begingroup$ If $(a,b)$ is an ordered pair, then $(a,(b,(c,d)))$ can be understood as an ordered 4-tuple. Tuples are useful in a lot of places, mostly when the sequence is not long enough to warrant indexing, e.g. see the definition of finite automaton. $\endgroup$ – dtldarek May 15 '15 at 11:51
  • 2
    $\begingroup$ You define $(a,b)$ as an object, but $(a,b)$ also appears in the definition. You probably meant $(a,b)=\{\{a\},\{a,b\}\}$ which is the Kuratowski definition of an ordered pair. $\endgroup$ – Asaf Karagila May 15 '15 at 12:35
2
$\begingroup$

As others have mentioned, it's typical to define $$(a_1,\ldots, a_n)=((a_1,a_2,\ldots, a_{n-1}),a_n)$$ though I like to call them "ordered $n$-tuplets" to distinguish from the alternative definition I give below (whose naming convention is rather standard when used for infinite sets, in which the above definition cannot be extended).

I think it's worth noting that there are a lot of different ways to extend the notion of an ordered-pair. For example, once ordered-pairs and ordered-triplets have been defined, you can define $A\times B$ for two sets $A$ and $B$ and a function as an ordered-triple $(A,B,f)$, where $f\subset A\times B$ is well-defined and left-total. For there, you can then "redefine" $n$-tuples to be surjective maps $f: \{0,\ldots, n-1\}\rightarrow A$. (Strictly speaking, we could get away with only having ordered-pairs if we want to just define an $n$-tuple to be a subset of $\{0,\ldots, n-1\}\times A$ for some set $A$ that is well-defined and left-total. But defining functions as triples is useful because it allows one to talk about surjectivity.)

Another approach is defining $(a,b):=\{a,\{a,b\}\}$, though that this satisfies the property that $(a,b)=(c,d)$ implies $a=c,b=d$ requires the Axiom of Foundation.

For the sake of differentiating between the ordered $n$-tuplets as defined above and these $n$-tuplets, I'll denote an $n$-tuple as $\langle a_0,\ldots, a_{n-1}\rangle$ where $a_0,\ldots, a_{n-1}$ are the images of the function $f$ on $0,\ldots, n-1$, respectively. The nice property that $n$-tuples have that $n$-tuplets don't have is that given an $n$-tuple $\mathbf{a}=\langle a_0,\ldots, a_{n-1}\rangle$ and an $m$-tuple $\mathbf{b}=\langle b_0,\ldots, b_{m-1}\rangle$, we have $\mathbf{a}=\mathbf{b}$ if and only if $n=m$ and $a_i=b_i$ for each $i\in\{0,\ldots, n-1\}=\{0,\ldots, m-1\}$, the reason being that two functions being equal implies their domains are equal.

As mentioned above, when we want to extend the notion of an $n$-tuple(t) to deal with indexing things by arbitrary sets, we have to use the notion of an $n$-tuple, defining an $I$-tuple (or a family of sets indexed by $I$) to be a surjective map $f:I\rightarrow A$. If want to regard these $I$-tuples as simply being elements of some infinite Cartesian product $\prod_{i\in I}{A_i}$ (where $(A_i)_{i\in I}$ is a family of sets indexed by $I$), we could choose the elements to either be indexed families $f:I\rightarrow A$ where $A\subset \bigcup_{i\in I}{A_i}$ and $f(i)\in A_i$ for each $i\in I$ (so $f$ is necessarily surjective by definition), or we could choose to take the elements to be simply functions $f:I\rightarrow \bigcup_{i\in I}{A_i}$ where $f(i)\in A_i$ for each $i\in I$. Restricting to the finite case where $I=\{0,\ldots, n-1\}$, we then have a few different ways we could have defined, say, $A\times B$ (though we did use the Kuratowski definition to define a function in the first place, nothing stops us from then considering as "ordered pairs" $2$-tuples rather than $2$-tuplets from now on).

Given these alternatives exist for definitions of $n$-tuple(t)s or infinite cartesian products, a question might be which is better. Ultimately, it's kind of an arbitrary choice, in the sense that there is a natural way to switch between the two that "preserves" how the canonical projections act on the Cartesian products.

$\endgroup$
0
$\begingroup$

We define an ordered pair by the definition $$ (a,b)=\{\{a\},\{a,b\}\} $$ You can check that this definition is valid for our purposes.
Now we define $n$- tuples inductivelly by $$ (a_1,a_2,\dotsc,a_n)=((a_1,a_2,\dotsc,a_{n-1}),a_n) $$

$\endgroup$
  • $\begingroup$ Actually this definition is valid in the presence of the axiom of regularity, but might not be valid if the axiom fails. $\endgroup$ – Asaf Karagila May 15 '15 at 12:28
  • $\begingroup$ I assumed that OP accepted the axiom of regularity... $\endgroup$ – k1.M May 15 '15 at 12:31
  • $\begingroup$ Since the axiom is not "naive" (in the sense that you don't really run into it when doing naive set theory) it's usually a good idea to bring it up when it is needed. Moreover note that the OP wrote Kuratowski's definition, which is not quite the one you suggest (or what the OP wrote, actually). $\endgroup$ – Asaf Karagila May 15 '15 at 12:35

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.