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I was studying the formal construction of numbers from $\mathrm{ZF}$, assuming the elements of the naturals being $\{\}, \{\{\}\}$, etc. Given the naturals, its ring properties and arithmetic, we can define the integers being a set of equivalence classes on $\Bbb N^2$. This logic is used from the integers to the complex numbers, so the tear always have sets of equivalence classes. My doubt is: since, for example, $\bigcup \Bbb Z = \Bbb N^2$, isn’t it wrong to formally say that $1_{\Bbb N} = 1_{\Bbb Z}$, that is, $1 \in \Bbb Z$? Since that the only thing we can assume is that there is a bijective map $\Bbb N \mapsto \Bbb Z$ (in fact, they’re both countable).

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marked as duplicate by Asaf Karagila real-analysis Feb 22 '18 at 6:23

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    $\begingroup$ Well, "formally right" is what you do not define. There is a subset of $\mathbb Z$ canonically identified with $\mathbb N$. For mathematicians other than specialists in foundations, we do, indeed, say $\mathbb N \subset \mathbb Z$. And similarly $\mathbb Z \subset \mathbb R \subset \mathbb C$. $\endgroup$ – GEdgar Feb 22 '18 at 1:01
  • $\begingroup$ Formally right: is a tautology from the axiomatic definition. $\endgroup$ – Lucas Henrique Feb 22 '18 at 1:12
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    $\begingroup$ While not exactly about the same sets, it's essentially the same question. See my answer there for my take on the question. $\endgroup$ – Asaf Karagila Feb 22 '18 at 6:24
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You are more or less correct. Starting with, say, the the von Neumann definition of the set of natural numbers, $\mathbb N$, if we define $\mathbb Z$ as equivalence classes of pairs of natural numbers, then $\mathbb N\not\subseteq\mathbb Z$. In fact, $\mathbb N\cap\mathbb Z = \varnothing$. Every element of $\mathbb Z$ is an infinite set, while no elements of $\mathbb N$ are. What is true, however, is that we have an injective function $\mathbb N\to\mathbb Z$ which sends $n_{\mathbb N}$ to $n_{\mathbb Z}$. We can then implicitly apply this injective function whenever we want to think of a natural number as an integer. Alternatively, we can redefine $\mathbb N$ to be the image of that injective function, and thus get $\mathbb N\subseteq\mathbb Z$ (though we'll have to use another name, $\omega$ say, for the von Neumann set of natural numbers).

In category theory, the very notion of "subobject" is an (equivalence class of) monomorphisms, which, in the case of the category of sets, are injective functions. To a categorist, "$\mathbb N\subseteq\mathbb Z$" means "there exists an injective function $\mathbb N\to\mathbb Z$", but in category theory you don't talk about the "elements" of an object.

Typically, we take a more abstract, "axiomatic" view of these objects when we're working with them so that the precise definitions we used don't matter. For example, $\mathbb Z$ is the initial object in the category of rings (or the free ring generated from the empty set). Many algebraic properties of the integers follow from just this. The construction of the integers as equivalence classes of pairs of naturals just shows that an object satisfying the above characterization actually exists. A different constructions of the integers would also work. It would produce a different but isomorphic ring. The categorical notion of a universal property often captures the abstract "interface" that we want to work with, while a set-theoretic construction may be necessary to prove that something exists that actually satisfies that "interface". Objects characterized by universal properties are, in general, only unique up to unique isomorphism, but that's more than enough typically (i.e. we want to treat isomorphic objects as "equal").

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