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In chapter 4 of Analysis I by Terence Tao, we have the following note about the set theoretic construction of the integers:

In the language of set theory, what we are doing here is starting with the space N × N of ordered pairs (a, b) of natural numbers. Then we place an equivalence relation ∼ on these pairs by declaring (a, b) ∼ (c, d) iff a + d = c + b. The set-theoretic interpretation of the symbol a — b is that it is the space of all pairs equivalent to (a, b): a — b := {(c, d) ∈ N × N : (a, b) ∼ (c, d)}; the existence of the set Z = {a — b : (a, b) ∈ N × N} of integers then follows from two applications of the axiom of replacement. However, this interpretation plays no role in how we manipulate the integers and we will not refer to it again. A similar set-theoretic interpretation can be given to the construction of the rational numbers later in this chapter, or the real numbers in the next chapter.

What caught my eye in this note is that Tao suggests we need to apply the Axiom of Replacement twice in order to justify the existence of Z, but Z may be constructed with a single application of replacement: for every (a, b) ∈ N × N, we replace it by a — b and thus get Z. This reasoning seems plausible, but something still feels off about it. Is this proof really correct, and hence a simplification of Tao's idea? Or is there some error in the logic behind it?

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    $\begingroup$ Does it matter if one uses an axiom twice in an argument instead of only once? $\endgroup$ Commented Aug 15, 2023 at 0:14
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    $\begingroup$ I think you are right about it not changing the nature of the proof, but as a beginner in writing proofs I think finding and checking this kind of simplifications helps a lot. $\endgroup$
    – HJE
    Commented Aug 15, 2023 at 0:25
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    $\begingroup$ It looks to me like he has in mind one use to define a-b, and a second use to define Z? $\endgroup$ Commented Aug 15, 2023 at 1:01
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    $\begingroup$ @JoelDavidHamkins Given that the rest of Z(F) is finitely axiomatized, I could see being interested in which (or how many) instances of the Separation/Replacement schemes is required. $\endgroup$ Commented Aug 15, 2023 at 11:38
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    $\begingroup$ @MikhailKatz It may not be an oversight of Tao, if he is simply arguing in ZFC, which is often axiomatized with replacement but not separation (since separation follows from replacement). In that case, he would be accounting correctly. The point of my answer is that if one is seeking to minimize the instances of replacement (as in the OP), then one may eliminate it entirely by falling back to the weaker Zermelo theory, which has separation but not replacement. In any case, these issues are usually covered in an intro undergraduate set theory class, so I don't regard them as research level. $\endgroup$ Commented Aug 15, 2023 at 13:03

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In fact one doesn't need the replacement axiom at all in order to implement this set-theoretic construction of the integers. The entire construction can be undertaken in Zermelo set theory, which lacks the replacement axiom.

Specifically, for a fixed copy of the natural numbers $\mathbb{N}$, one can form the set of pairs using applications of the power set axiom and mere separation, and then define the set of equivalence classes using another application of power set and separation. Indeed, by applying first a suitable number of power set operations, with instances of union, one can apply a single application of the separation axiom to arrive at the desired copy of $\mathbb{Z}$ realized as equivalence classes of the same-difference equivalence relation.

The larger lesson is that the replacement axiom is needed in set theory generally only for associations mapping the elements of a set into the universe in such a way that there is no previously known upper bound of a set from which those associated individuals might arise. This reaching-upward aspect is what gives the power of ZFC set theory over Zermelo's strictly weaker theory Z without replacement. It is consistent with Zermelo's theory, for example, that $\aleph_\omega$ does not exist, even when $\omega$ exists and we can define the map $n\mapsto\aleph_n$. One needs replacement to show that the cardinals do not end prematurely in this way, as in Zermelo set theory there is no way to provide a bounding set for the $\aleph_n$'s.

But when one already has an upper bound for the witnessing objects, as is the case here since the equivalence classes are sets of pairs, and the set of all equivalence classes is thus a set of sets of pairs, then we can use merely the separation axiom to get at what we want.

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  • $\begingroup$ This is quite interesting. So as the construction of the integers depends only in weaker notations found in Zermelo's theory, one can have multiple proofs that build Z using zero, one or more instances of the Axiom of Replacement? $\endgroup$
    – HJE
    Commented Aug 16, 2023 at 0:33

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