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In the book The Foundations of Mathematics by Kenneth Kunen, in a chapter about set theory formalizations, the following first order logic sentence is provided (referred to as the Axiom of Comprehension/Separation Schema):

$\forall z \Big (\exists y \color{red}{\forall x} \big ( \color{red}{x} \in y \leftrightarrow x \in z \land \varphi(\color{red}{x}) \big ) \Big )$. I have colored in $\color{red}{\text{red}}$ the variable that I will be focusing on for the remainder of this question.

Consider the following two sets:

$A = \{ x \in \mathbb N \ |\ x \lt 5\}$.

$B = \{ \langle s, t \rangle \in \mathbb N \times \mathbb N \ |\ s \lt 5 \land t\lt 4 \}$

Now, let's say I wanted to look at set $A$ and confirm that it exists through the perspective of the Axiom of Comprehension Schema:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \land x \lt 5)$ where $\varphi(x):= x \lt 5$. Cool. Set A exists.

To confirm set $B$'s existence, however, I run into an issue when it comes to inserting the predicate $\varphi$:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \times \mathbb N \ \land \text{something} )$

I use the phrase "$\text {something}$" because "$\varphi (x)$" does not appear to decompose correctly into a statement about two variables...i.e. $\varphi(x)$ cannot be used to represent "$s \lt 5 \land t \lt 4$" because $s$ and $t$ are not necessarily equal, and therefore do not represent one object $x$.

A quick fix seems to be to switch $\forall x$ with $\forall s \forall t$ in order to generate the following statement:

$\exists y \forall s \forall t (\langle s,t \rangle \in y \ \leftrightarrow \ \langle s,t \rangle \in \mathbb N \times \mathbb N \ \land s \lt 5 \land t \lt 4 )$

I am uncertain why this is something that is allowed. Specifically, this seems to suggest that $\forall x \ \varphi(x)$ is 'semantically' (regular English definition) equivalent to $\forall s \forall t \ \varphi (s,t)$.

From what I understand about first order logic, $\forall x \ \varphi (x)$ essentially means "For all singular objects in the domain of discourse, the following sentence about a singular object is true". Comparatively, I interpret $\forall s \forall t \ \varphi(s,t)$ as saying "For any two objects in the domain of discourse, the following sentence is true about two objects".

Is the syntax "$\ s,t\ $" a singular object in the domain of discourse? It doesn't strike me as one...hence my confusion.

Any input is greatly appreciated! Thank you.

S.C.
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  • I think I have found a workaround to show that $B$ is a set, but I am unsure about your core claim of replacing $\forall x$ with $\forall s\forall t$ – supinf Dec 02 '20 at 20:37

3 Answers3

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You use the projection functions - or rather, the projection relations. These are the binary relations $\pi_0(x,y)$ and $\pi_1(x,y)$, which we read as "$x$ is an ordered pair with left coordinate $y$" and "$x$ is an ordered pair with right coordinate $y$." Note that the expression "$u=\langle v,w\rangle$" is really shorthand for "$\pi_0(u,v)\wedge\pi_1(u,w)$."

Writing out the projection relations in terms of $\in$ alone is a good exercise. Once you've done this, you can produce $B$ by applying Separation to the formula

$$\varphi(x)\equiv \forall s,t(\pi_0(x,s)\wedge \pi_1(x,t)\rightarrow s<5\wedge t<4).$$

Separation applied to this formula, with "container set" $z=\mathbb{N}\times \mathbb{N}$, says that there is a set $B$ consisting of exactly those $a$ which are elements of $z$ and which satisfy $\varphi$ - and this $B$ is precisely what we want.

Noah Schweber
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  • If you recall this post I made a while back (which you responded to) --> https://math.stackexchange.com/questions/3855153/method-explanation-for-proof-used-in-the-following-proposition-if-sz-z-tz-z, there was an issue of interpreting the universal quantifier of a single object as the universal quantifier of two objects. In particular, in one of your comments under you answer, you wrote, "But the universal quantifier over $v$ really is over $x$ and $y$ as well - they're introduced as components of $v$". Is this a similar concept? Or unrelated? – S.C. Dec 02 '20 at 20:46
  • @S.Cramer Yup, if memory serves (it's been a while) this is exactly what's implicitly going on there: we can quantify over (say) pairs by quantifying over individual elements and then using projections. – Noah Schweber Dec 02 '20 at 20:47
  • B-e-a-utiful. Thank you! Let me digest this answer for a little bit now. Cheers~ – S.C. Dec 02 '20 at 20:48
  • For clarity, your sentence "...$\pi_0(x,y)$ and $\pi_1(x,y)$, which we read as '$x$ is an ordered pair with left coordinate $y$' and '$x$ is an ordered pair with right coordinate $y$' " is specific to MY question, yes? More generally, we would say, for example, $\pi_0(x,y)$ is read as, "$x$ is an element of a cartesian product and $y$ is the $0$th index (or 'most far left component') of that element". Is that correct? – S.C. Dec 02 '20 at 21:00
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    @S.Cramer Sure - although it's not really established notation (it's usually clear from context though). Really, once you're looking beyond pairs you should be thinking about tuples as functions from an indexing set (recall that these are themselves sets of ordered pairs). At that point, we're just talking about function evaluation. This is necessary to say e.g. "every coordinate of this tuple is nonzero" or similar - if we introduce a separate formula for each coordinate, things get messy (or impossible!). See also the discussion here. – Noah Schweber Dec 02 '20 at 21:06
  • I thought about your comment ("Writing out the projection relations in terms of $\in$ alone is a good exercise") and came up with something that I hoped you might direct me on. I interpreted that recommendation of yours as saying, "Use set theory FOL to describe what exactly $\pi_{n \in \mathbb N} (x,y)$ means". There are 3 important things to note: 1) $x$ needs to be a function (i.e. $x={\langle 0,a \rangle , \langle 0,b \rangle ... }$ would not qualify, 2) $x$'s domain needs to be the subset of $\mathbb N$, and 3) encode the necessary property of $y$ as it relates to the subscript $n$ – S.C. Dec 07 '20 at 09:07
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    Using $\pi_0$ as our example: $\langle x,y \rangle \in \pi_0 \iff \text{dom}(x) \subseteq \mathbb N \land \forall z \big (z \in \text{dom}(x) \rightarrow \exists ! y \varphi(x,y) \big) \land \exists w \big (w \in x \land \forall z (z\in w \rightarrow 0 \in z) \land (\bigcup w \setminus 0 = \emptyset \rightarrow y=0) \land (\bigcup w \setminus 0 \neq \emptyset \rightarrow y \in \bigcup w \setminus 0) \big )$ – S.C. Dec 07 '20 at 09:14
  • For the above expression, I used the Kuratwoski implementation of an ordered pair, which is why some of those statements are written the way that they are – S.C. Dec 07 '20 at 09:20
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Here is a possible workaround in order to show that $B$ is a set:

You could use a single variable $w$ (to encode the pair $\langle s,t\rangle$) and then use $$\phi(w):=\exists s\exists t\;\, w=\,\langle s,t\rangle \land s<5 \land t<4$$

supinf
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    Note to the OP that this is the same as my answer, but snappier: "$w=\langle s,t\rangle$" is shorthand for "$\pi_0(w,s)\wedge\pi_1(w,t)$" for appropriate projection formulas $\pi_0,\pi_1$. – Noah Schweber Dec 02 '20 at 20:45
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You can choose $\varphi(x)$ to be $x_0<5\land x_1<4$, if you have the notation $x=\langle x_0,\,x_1\rangle$ for $2$-tuples. Mathematicians often abbreviate this by saying $\varphi(\langle s,\,t\rangle)$ iff $s<5\land t<4$, with it being taken as implicit that $\neg\varphi(x)$ whenever $x$ isn't a $2$-tuple (although that wouldn't matter in the present context).

However, the existence of a Cartesian product $X\times Y$, such as $\Bbb N\times\Bbb N$, is proven with several axioms:

  • First we convince ourselves ordered pairs exist, e.g. on Kuratowski's definition with unordered-pair (an axiom in Zermelo set theory, a theorem in ZF) repeatedly.
  • Replacement tells us first that $X\times\{a\}$ exists for each $a\in Y$, then that there is a set of these Cartesian products.
  • We can now take a union to get $X\times Y$.

If this question presupposes we already accept the above, we can prove $B$ exists by writing $B=X\times Y$ with$$X=\{s\in\Bbb N|s<5\},\,Y=\{t\in\Bbb N|t<4\},$$which lets us use comprehension twice, with very obvious choices of $\varphi$.

J.G.
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