In ZFC set theory, every element object is a set. We know or feel that if $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set.

My question is: Why if $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set?

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    $\begingroup$ By definition, the Cartesian product of two sets is the SET of all ordered pairs of blablabla $\endgroup$ – Dante Nov 2 '16 at 1:26
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    $\begingroup$ @Dante That's not the point of this question. $\endgroup$ – Stefan Mesken Nov 2 '16 at 1:40

tldr; $X\times Y$ is a definable subset of $\mathcal{P}(\mathcal{P}(X \cup Y))$ (assuming that you're using, say, the Kuratowski definition of an ordered pair).

More precisely, suppose $(x,y) = \{\{x\},\{x,y\}\}$. Given the elements $x \in X, y \in Y$, the sets $\{x\}$ and $\{x,y\}$ are in $\mathcal{P}(X \cup Y)$, so the set $\{\{x\},\{x,y\}\} \in \mathcal{P}(\mathcal{P}(X \cup Y))$. Then define $X\times Y$ to be the set of all sets $z$ in $\mathcal{P}(\mathcal{P}(X \cup Y))$ such that there are $x,y$ with $z=(x,y)$. You need to do a little bit more than this to actually create a wff in the first-order language of set theory, but not that much more.

For a full wff, we have the following: $$\varphi(z) = \exists x\exists y\exists v\exists w(x\in X \wedge y\in Y \wedge \mathrm{pair}(v,x,x) \wedge \mathrm{pair}(v,x,y) \wedge \mathrm{pair}(z,v,w))$$ where $$\mathrm{pair}(z,x,y) = \forall w (w\in z \leftrightarrow ((w=x)\vee (w=y))$$ In other words, $\mathrm{pair}$ is a wff with three free variables which expresses the fact that $z=\{x,y\}$. $\varphi$ expresses that $z$ is itself a pair of elements $v,w$ which are respectively, $v=\{x,x\}=\{x\}$ and $w=\{x,y\}$. In other words, $z=\{\{x\},\{x,y\}\}$.

Then $$X\times Y = \{ z \in \mathcal{P}(\mathcal{P}(X\cup Y) \mid \varphi(z)\}$$ (note that $\varphi$ has the parameters $X,Y$).

This uses the Axiom of Power Set, Axiom of Pairing, Axiom of Union, and the Axiom Schema of Restricted Comprehension.

Why didn't we just say $$X\times Y = \{ (x,y) \in \mathcal{P}(\mathcal{P}(X\cup Y)) \mid x\in X \wedge y\in Y\}$$ ?

Well, recall what the Axiom Schema of Restricted Comprehension actually says: for every set $X$ and every wff $\varphi(x)$ with the free variable $x$, there exists a set $Y$ whose elements are exactly those in $x\in X$ which satisfy $\varphi(x)$. But the aforementioned 'definition' isn't in that form!

How do we normally get away with this, though? Certainly we say things like $\{f(x)\in Y \mid x\in X\}$ for the image of a function $f$! This uses another axiom of $\mathsf{ZFC}$, called the Axiom Schema of Replacement. As shown above, we don't need to use it, and judging from the answer you had written up, it seems you haven't gotten to Replacement yet.

  • $\begingroup$ @Stefan thanks for the edit, for some reason when writing I was just assuming I was showing it for $X\times X$. $\endgroup$ – Hayden Nov 2 '16 at 1:46
  • $\begingroup$ I wasn't sure whether I should give you a heads up about this edit. But since I was certain about what you meant to say, I thought it would be fine. Glad to see that this seems to be the case! $\endgroup$ – Stefan Mesken Nov 2 '16 at 1:47

Proof. Since $x\in X$, we have $x\in X\cup Y$. Similarly, $y\in X\cap Y$.It follows that the sets {$x$} and {$x,y$} are subsets of $X\cup Y$ and hence elements of $\mathcal{P}(X\cup Y)$. Therefore, {{$x$},{$x,y$}} is a subset of $\mathcal{P}(X\cup Y)$. Hence, $(x,y)=${{$x$},{$x,y$}}$\in \mathcal{P}(\mathcal{P}(X\cup Y))$ . Thus all the elements of $X\times Y$ are elements of $\mathcal{P}(\mathcal{P}(X\cup Y))$. Now among the elements of $\mathcal{P}(\mathcal{P}(X\cup Y))$ we will distinguish the ones of the form $(x,y)$ by a formula $\mathcal{P} (z)$, i.e., by a property. In other words, we will find a formula $\mathcal{P} (z)$ such that for all $z$, the following will hold:

$\mathcal{P} (z) \Leftrightarrow \exists x\exists y(x\in X\wedge y\in Y\wedge z=${{$x$},{$x,y$}$)$.

If we can do this, then we will have $X\times Y=$ {$z\in\mathcal{P}(\mathcal{P}(X\cup Y)):\mathcal{P}(z)$}, and so by A3, $X\times Y$ will be a set.

  • $\begingroup$ Can you check my last proof? $\endgroup$ – PozcuKushimotoStreet Nov 2 '16 at 16:01
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    $\begingroup$ You shouldn't just create a new answer for each proof attempt; edit your last answer or delete that one. The only issue with the proof now is that, strictly-speaking, in the language of set theory you cannot write $z=\{\{x\},\{x,y\}\}$ in a wff; $\{\{x\},\{x,y\}\}$ is not even in the syntax of set theory. That being said, many people will ignore this, with the understanding that $z=\{\{x\},\{x,y\}\}$ is representative of a wff which expresses that $z$ has the indicated form. If this is homework, you should ask for clarification from your professor about what their expectation is. $\endgroup$ – Hayden Nov 2 '16 at 18:14

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