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I've recently learned that unlike in naive set theory, $N\not\subset Z\not\subset Q\not\subset R$ in ZFC.

Here are some discussions on it:

I was wondering if there were any widely-accepted and formal set theories in which $N\subset Z\subset Q\subset R$, just as our intuition tells us.

SMMH
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    You can define the reals axiomatically, then extract from that the naturals, or the rationals, or the integers, or whatever. https://math.stackexchange.com/a/2437917/ – Xander Henderson Aug 12 '23 at 21:44
  • Well, $\Bbb N\subseteq\Bbb Z$ is literally true if you just take a particular model of $\Bbb Z$ that contains another particular model of $\Bbb N$. Because remember we only care about these objects in terms of their abstract behaviour than what their literal elements are. And all of that works in ZFC. $\Bbb N\not\subset\Bbb Z$ is not a symptom of ZFC, it's a symptom of the canonical ways in which $\Bbb N$ and $\Bbb Z$ are constructed. But equally one could construct $\Bbb Z$ in the canonical way and define $\Bbb N$ to be its positive subset. Whatever – FShrike Aug 12 '23 at 23:07
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    This is not a question of the set theory, but rather of the definitions we make in the set theory. ZFC on its own says nothing about these number systems - we have to define them. According to the standard constructions / definitions, these number systems are not contained in each other, but it's easy to modify the definitions (without modifying ZFC in any way!) so that they are contained in each other. – Alex Kruckman Aug 13 '23 at 03:08
  • @XanderHenderson what if we also have complex numbers? and quaternions? and octonions? and...? There is really no mother set that contains all of these, so I think defining the smaller ones based on the bigger ones is not going to work out. We have to do the opposite and define bigger ones based on the smaller ones. – SMMH Aug 15 '23 at 15:14
  • @FShrike what about $\mathbb{Z}\subset\mathbb{Q}$? what about $\mathbb{N}\subset\mathbb{R}$? – SMMH Aug 15 '23 at 15:19
  • @AlexKruckman could you please give a link to those "standard constructions / definitions"? is there a reasonable consensus on them in the math community? – SMMH Aug 15 '23 at 15:20
  • @SMMH Yes, there is a reasonable consensus. Open up any introductory set theory textbook. Introduction to Set Theory by Hrabacek and Jech is a good choice. Or Googling "set theoretic construction of number systems" gives a number of references. – Alex Kruckman Aug 15 '23 at 15:24
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    Regarding your comment to @XanderHenderson: Let's say you have a construction of $\mathbb{N}$ and a construction of $\mathbb{Z}$ according to which $\mathbb{N}\not\subseteq \mathbb{Z}$, and you're dissatisfied with this situation. Well, we have an injective function $i\colon \mathbb{N}\to \mathbb{Z}$ mapping each natural number to "its copy" in the integers. Now you're free to define $Z' = (\mathbb{Z}\setminus i(\mathbb{N})) \cup \mathbb{N}$ (that is, remove the "copies" and replace them with the "real" natural numbers) and call $Z'$ the "integers" instead of $\mathbb{Z}$. – Alex Kruckman Aug 15 '23 at 15:29
  • @AlexKruckman Which is, indeed, exactly what I suggested doing in the answer I linked: "Alternatively, instead of declaring the existence of the reals by fiat (item 5, above), we could build them up from scratch via the usual process (i.e. build the integers as equivalence classes of naturals, then the rationals as equivalence classes of integers, then the reals as Dedekind cuts). This construction of the reals will satisfy the above axioms. Then, you can build the real natural numbers via the ["inductive set" definition in the question]." ;) – Xander Henderson Aug 15 '23 at 15:31
  • It makes no difference if you do this, since there is a canonical bijection $\mathbb{Z}\to Z'$, so anything "structural" (i.e., not looking under the hood at what set happens to represent what number) that you can prove about $\mathbb{Z}$, you can also prove about $Z'$. Now there was nothing special about $\mathbb{N}$ and $\mathbb{Z}$ here. If you do this systematically as you define each larger number system, you can obtain $\mathbb{N}\subset \mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}\subset\dots$ – Alex Kruckman Aug 15 '23 at 15:31
  • @XanderHenderson Oh yes, I see. I didn't click though, sorry! – Alex Kruckman Aug 15 '23 at 15:32
  • @AlexKruckman No, I probably rambled on too long, and didn't make that point obvious enough. :D – Xander Henderson Aug 15 '23 at 15:33

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