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Ok, so this question seems obvious, right?

But what I mean is the numbers in the way they are logically / axiomaticaly defined in the foundations of Mathematics. As far as I know, the naturals are in most cases defined as sets, I.e. zero being the empty set, one the set which contains the empty set, and so on. The real numbers however are defined as equivalence classes of Cauchy-sequences or as Dedekind cuts... So how can the natural number 3, which is a set, be an an element of a set which contains only equivalence classes? Same for the rationals, they are equivalence classes as well, right?

So wouldn't you technically be forced to say that the naturals are not a subset of the rationals or the reals?

Of course you can embedd them; but I do not unterstand how the natural number 3 can be really identical with the real number 3.

P. Grewe
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    There is a subset of the real numbers isomorphic to the naturals. It is convenient sometimes to treat the isomorphism as an inclusion, so we can refer to the real number $n$ when $n\in\mathbb N.$ A real stickler would either refer to as $i(n)$ or make a very complicated alternate set of definitions for $\mathbb Z,\mathbb Q,$ and finally $\mathbb R$ so that the maps are set inclusions. There is no real harm in treating these as inclusions, without the awkward definitions. – Thomas Andrews Nov 15 '21 at 23:25
  • I feel that my answer to another question may address your question. The short version is "Yes", if you start with the real and define the naturals as a an inductive set containing 0 and 1. But if you build the reals out of Cauchy sequences or Dedekind cuts, then the answer is "No", but there is a subset of the reals which is isomorphic to the naturals. It depends on your definitions. – Xander Henderson Nov 15 '21 at 23:26
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    All of the “inclusions” $$\mathbb N\subset \mathbb Z\subset \mathbb Q\subset \mathbb R\subset\mathbb C$$ are technically not true, as we normally define these sets. It is worth noting early on why we can treat them as true. – Thomas Andrews Nov 15 '21 at 23:31
  • An interesting example is $\mathbb C$ contained in the quaternions. Here, there isn’t one “natural” inclusion, so, while the quaternions are an extension of $\mathbb C,$ we don’t necessarily think of $\mathbb C$ as a subset of $\mathbb H.$ – Thomas Andrews Nov 15 '21 at 23:37
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    Do you not believe that the natural numbers existed before a model of them was made in a set theory? Same question about the reals? Then, why would it matter how they're modeled to answer a question of who includes whom? – Eric Towers Nov 16 '21 at 00:06
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    Interestingly, there are lots of inclusions $\mathbb N\to\mathbb Z$ that are isomorphisms as ordered sets. There is only one that is an isomorphism under addition. There are lots of inclusions $\mathbb Z\to\mathbb Q$ which are isomorphisms under addition, but only one also an isomorphism under multiplication. There are also (if memory serves) loads of algebraic inclusions $\mathbb R\to\mathbb C,$ but only one continuous such inclusion. I think the only case where there is only one way using the prior level’s condition is $\mathbb Q\to\mathbb R.$ – Thomas Andrews Nov 16 '21 at 00:17
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    The issue is with that little word... "really". What are numbers, "really"? We have different theories "speaking of" numbers: set theory, real analysis (with e.g. Dedekind cuts construction), axiom for fields: all of them "describes" mathematical structures with the same properties, i.e. they all "look the same". – Mauro ALLEGRANZA Nov 30 '21 at 15:28

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