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I have been watching the YouTube series 'Start Learning Mathematics' by The Bright Side of Mathematics. I am currently on episode #3 of the set series and he's just introduced us to 'ordered pairs.' In more detail, the episode is about the Cartesian product, and how it is the set of all ordered pairs of sets A & B (at least this is what I understood from it). He gives the proper definition: $$A\times B:= \{(a,b)\mid a\in A\ \wedge\ b\in B\}$$

And then he defines the ordered pair through Kuratowski's method/proof. He starts by saying For elements $x,y$ write: $$(x,y) := \{\{a\},\{a,b\}\}$$ and $$(x,y)=(\tilde{x},\tilde{y}) \Leftrightarrow \{x\}=\{\tilde{x}\}\ \wedge\ \{y\}=\{\tilde{y}\}$$ $$\Leftrightarrow x = \tilde{x}\ \wedge\ y = \tilde{y}$$

It has left me very confused. I mostly understand the last bit, it seems obvious that if you have points $(a,b) = (c,d)$ then $a = c$ and $b = d$, but I still don't understand why the first set has $\{a,b\}$ in it and not just $\{b\}$, and what the entire point of these are.

I would appreciate any help, thank you!

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    $\begingroup$ The problem with $(a,b) = \{\{a\}, \{b\}\}$ is that it would also mean $(a,b)=(b,a)$ and we were trying to ensure the pair was ordered. $\endgroup$ Commented May 29 at 16:29
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    $\begingroup$ "It seems obvious that if you have points $(a,b)=(c,d)$ then $a=c$ and $b=d$" If ordered pairs were not defined in an adequate fashion then this might not have been true. For example, we do have that $\{a,b\}=\{c,d\}$ but this does not imply that $a=c$... for instance when $a=d=1$ and $b=c=2$ we have the unordered set $\{1,2\}$ is equal to the unordered set $\{2,1\}$ despite $a\neq c$. $\endgroup$
    – JMoravitz
    Commented May 29 at 16:39
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    $\begingroup$ The punchline is that we wanted a way to define ordered pairs so that they have the properties we want them to have. Kuratowski's definition does exactly that and does so in a way that builds off of already existing notions of set theory that do not require the creation of new axioms. $\endgroup$
    – JMoravitz
    Commented May 29 at 16:43
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    $\begingroup$ The stated purpose of Kuratowski's formalisation is that we have $\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}$ iff $a=c$ and $b=d$. We could also have $(a,b):=\{\{0,a\},\{1,b\}\}=\{\{\{\},a\},\{\{\{\}\},b\}\}$ etc. as you can check. These constructions have the familiar property of an ordered pair. From Munkres' Topology: "I think it is fair to say that most mathematicians think of an ordered pair as a primitive concept rather than thinking of it as a collection of sets!" $\endgroup$ Commented May 29 at 19:50
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    $\begingroup$ Note that mathematicians never use this in practice - it's basically just a way to get around adding an additional axiom about ordered pairs to ZFC. $\endgroup$ Commented Jun 1 at 0:25

2 Answers 2

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Here are some words that people really should use a lot more in mathematics than they do: there are two things that people mean when they talk about "defining" a mathematical object, which you might call specifications and constructions (as far as I know there is no widely adopted language for this distinction in mathematics). A specification is like a blueprint; you specify the properties you want that object to have. A construction is like actually building a house; you actually construct the object to demonstrate that it really exists.

The specification of "ordered pair" is: whatever ordered pairs are, they should take as input two objects $a, b$ and return as output a third object $(a, b)$, and this third object should have the property that $(a, b) = (c, d)$ iff $a = c$ and $b = d$. In other words an ordered pair should be what it sounds like: it should contain exactly the information of the two objects that make it up, in order.

Kuratowski's "definition" of an ordered pair should more precisely be called a construction; it constructs, in set theory, an object satisfying the above specification. As mentioned in the comments, $\{ \{ a \}, \{ b \} \}$ does not satisfy the above specification, because $\{ \{ a \}, \{ b \} \} = \{ \{ b \}, \{ a \} \}$; this instead is a possible construction of an unordered pair.

Other constructions of ordered pairs are possible; see Wikipedia.

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    $\begingroup$ Philosophically speaking you might argue that ordered pairs are exactly as primitive a concept as sets, if not more so. But this is what you do if you want to start with set theory and build everything else from there. $\endgroup$ Commented May 29 at 20:09
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    $\begingroup$ +1. Just commenting to mention the keywords "analytic" and "synthetic", which correspond to constructions and specifications respectively. See for example this discussion on MathOverflow: mathoverflow.net/questions/405011/… $\endgroup$ Commented May 30 at 1:01
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    $\begingroup$ Re: reducibility of the notion of "ordered pair". When Peano formulated the notion in $1987$ he wrote:. "The idea of a pair is fundamental, i.e., we do not know how to express it using the preceding symbols." The historical article by Kanamori that I cite there is an interesting read. $\endgroup$ Commented May 30 at 2:31
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    $\begingroup$ What we are really after is something that matches the specification, because that's what determines what you can actually use it for. The construction is there to be done once, mostly to show that it's actually possible to make something that follows the specification. A big part of the work in this problem lies in showing that the construction matches the specification. After that has been done, you basically forget the construction, and use only the specification every time you interact with an ordered pair ever again. This pattern crops up all over mathematics. $\endgroup$
    – Arthur
    Commented May 30 at 8:23
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    $\begingroup$ +1 The computer science (object oriented programming) analogue that might help some readers is the difference between the promise a method makes - the interface visible to the caller - and the implementation inside the object. $\endgroup$ Commented May 30 at 16:00
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There are two aspects. First is that the notion of "ordered pair" is not the same as Kuratowski's version of the ordered pair, which is what you are asking about. Like many other things, the core notion is an abstraction (or interface), which specifies how the relevant objects must behave. For ordered pairs, we want them to have the following behaviour:

$∀x,y,z,w\ ( \ ⟨x,y⟩ = ⟨z,w⟩ ⇔ x = z ∧ y = w \ )$.

That is all we want. So mathematicians are perfectly fine with absolutely nothing more than this, treating ordered pairs as a primitive notion that satisfies the above abstract property. Ordered pairs are so crucial to mathematics that no mathematician would ever accept a foundational system for mathematics that cannot support this basic notion. Hence it is very important to know whether we can implement ordered pairs using whatever primitive notions we have available in any foundational system that we would like to use!

In some type theories, ordered pairs are primitive notions, so we do not have to look for an implementation. But in most presentations of ZFC set theory, the only primitive notion is the membership relation $∈$, and so we need to demonstrate a way of implementing ordered pairs. One way is Kuratowski's definition. That doesn't imply that ordered pairs are as Kuratowski's implementation; of course the abstraction is not the implementation. Besides, we could just as well define $⟨a,b⟩$ as $\{\{a,b\},\{b\}\}$, or as $\{\{\{\},\{a\}\},\{\{b\}\}\}$.

The second aspect is that whatever you were looking at failed to explain why $⟨x,y⟩ = ⟨z,w⟩$ implies $\{x\} = \{z\}$ and $\{y\} = \{w\}$. This is not as trivial as you might think. You need to be careful not to assume that $\{x,y\}$ has two (distinct) members, because it may be that $x = y$. To do this properly, you can first prove that given any $p = \{\{x\},\{x,y\}\}$ we have that $x$ is the unique member of both members of $p$, and then we have two cases:

  • If $x = y$, then $p = \{\{y\}\}$, so for any $z$ such that $p = \{\{x\},\{x,z\}\}$ we have $\{y\} = \{x,z\}$ and hence $y = z$.
  • If $x ≠ y$, then $\{x,y\}$ has two (distinct) members unlike $\{x\}$, so for any $z$ such that $p = \{\{x\},\{x,z\}\}$ we have $\{x,y\} = \{x,z\}$ and hence $y = z$.

I have left out a few tiny steps, but I hope you can carefully fill them in.

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