3
$\begingroup$

I got this idea while reading a textbook. Previously I read that ordered pairs can be formalized in axiomatic set theory — for example, $(a, b) := \{\{a\}, \{a, b\}\}$ is viable solution. It has the property we would like to have, i.e. $ (a, b) = (c, d) $ iff $a = c$ and $b = d$.

However, this only deals with pairs. If the Cartesian product of two sets $X$ and $Y$ is defined as follows: $$ X\times Y := \{(x, y)\in\mathcal P(\mathcal P(X\cup Y))| x\in X \land y \in Y\}$$ Then, what does it mean by writing $\mathbb R^3$? If this refers to $\mathbb R \times \mathbb R \times \mathbb R $, which of these does it exactly express: $(\mathbb R \times \mathbb R) \times \mathbb R $ or $\mathbb R \times (\mathbb R \times \mathbb R)$? To make things worse, sometimes we write $(V^*)^p \times V^q$ to donate a $(p+q)$-tuple, such that the first $p$ elements are from the $V^*$, and the rest of them are from $V$. The formalization of ordered pairs and Cartesian products discussed above doesn’t seem to able to solve the problems directly.

Thus, I wonder whether the idea of ‘flattened’ ordered pairs may be useful: to make the ordered pairs satisfy yet another axiom: $\forall a, b, c(a,(b,c)) = ((a, b), c)$. Hence, the distinction between $(a,(b,c))$ and $(a,(b,c))$ is eliminated, and we can write uniformly: $(a,b,c)$.

To summarize: the theory of ‘flattened’ ordered pairs includes a primitive binary function $\mathscr P$ and two axioms: $$ \forall a \forall b \forall c \forall d ( \mathscr P ab \equiv \mathscr P cd \leftrightarrow a \equiv c \land b \equiv d)$$ $$ \forall a \forall b \forall c( \mathscr P \mathscr P abc \equiv \mathscr P a \mathscr P bc )$$

Does it has any model in the axiomatic set theory?

I tried $\langle a, b\rangle := \{\{a\}, \{a, b\}\}$ and $(a_1, \ldots, a_n):=\{\langle 1, a_1 \rangle, \ldots , \langle n, a_n\rangle\}$ which gives the $n$-tuple directly. I also thought about definition involving equivalent class. However, I still don’t know how to define the function $\mathscr P$, unless the if the parameter is a tuple is previously known. (Is this possible?)

Finally, my questions are:

  1. Is there a more elegant way to solve the problem of defining multiple cartesian products?

  2. Can we build the ‘flattened’ ordered pairs from the set theory?

$\endgroup$
9
  • 9
    $\begingroup$ Doesn't $(a,(b,c))=((a,b),c)$ imply $a=(a,b)$ for all $b$? And doesn't $(a,b)=a=(a,b')$ imply $b=b'$? So how could both your axioms hold in a universe with more than one element? $\endgroup$
    – bof
    Dec 10 '18 at 10:21
  • 4
    $\begingroup$ I do not see what is wrong with noting that there is a canonical bijection between $A\times(B\times C)$ and $(A\times B)\times C$, and then defining $A\times B\times C$ as whichever one you prefer. $\endgroup$
    – Servaes
    Dec 10 '18 at 10:23
  • 1
    $\begingroup$ @bof That seems like a fine answer to me. $\endgroup$
    – Servaes
    Dec 10 '18 at 10:25
  • 1
    $\begingroup$ @Servaes I think that I have got your idea, but I am still a little bit concerned. If ‘the same’ has become ‘canonically in bijection’, can we define a equivalent class over all the possible pair that are ‘canonically map to each other’ to return to the equality level? $\endgroup$
    – fantasie
    Dec 10 '18 at 11:58
  • 4
    $\begingroup$ Having used the standard definition of pairs to define functions and having defined the natural numbers, you can define $X_1 \times X_2 \times \ldots \times X_n$ to be the set of functions $i \mapsto x_i$ in $\{1, 2, \ldots n\} \to X_1 \cup X_2 \cup \ldots X_n$ such that $x_i \in X_i$ for $i = 1, 2,\ldots n$. That avoids an arbitrary choice of where to put the brackets in an iterated binary product and aligns nicely with how one writes vectors in $\Bbb{R}^n$, $\Bbb{C}^n$ etc. I think that's as close as you'll get to "returning to the equality level". $\endgroup$
    – Rob Arthan
    Dec 10 '18 at 20:59
1
$\begingroup$

A simple thing you can do is pick a fixed parenthesization: for example, you can choose to parenthesize starting on the right, so that e.g. $\mathbb{R}^4$ is $\mathbb{R} \times (\mathbb{R} \times (\mathbb{R} \times \mathbb{R}))$, or you can choose to parenthesize starting on the left, which is flipped. The problem is that with this definition the Cartesian product fails to be associative on the nose, so that for example $\mathbb{R}^2 \times \mathbb{R}^2 \neq \mathbb{R}^4$, and you might consider this unsatisfying.

What is true, however, is that the Cartesian product is monoidal. This means that even though it's not associative on the nose we can define a family of bijections

$$a_{X, Y, Z} : X \times (Y \times Z) \cong (X \times Y) \times Z$$

called associators which satisfy a somewhat complicated set of axioms, the point of which is to guarantee that any two ways of using associators to identify two parenthesizations of a Cartesian product agree.

Most binary operations on mathematical objects that you're familiar with fail to be associative on the nose (depending on how you construct them) but are still monoidal in this way, for example tensor products of vector spaces.

$\endgroup$

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.