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Usually, numbers sets are constructed from $\mathbb N$ (with Peano's axioms) : then $\mathbb Z$ is viewed as a quotient set of $\mathbb N\times\mathbb N$, $\mathbb Q$ as a quotient set of $\mathbb Z \times \mathbb Z$, and $\mathbb R$ is defined with Dedekind cuts or Cauchy sequences of rationals. Well, we can define $\mathbb C$ as $\mathbb R\times\mathbb R$ or $\mathbb R[X]/(1+X^2)$.

These sets are linked through injective morphism, i.e. $\mathbb N \hookrightarrow \mathbb Z \hookrightarrow \mathbb Q \hookrightarrow \mathbb R \hookrightarrow \mathbb C$. However, there is no inclusion like $\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$ because, there are different sets (by construction)...

How to define numbers sets as being subsets of next one ?

Does it possible, to construct $\mathbf N$, $\mathbf Z$, $\mathbf Q$, $\mathbf R$ and $\mathbf C$ instead of $\mathbb N$, $\mathbb Z$, $\mathbb Q$, $\mathbb R$ and $\mathbb C$ then create some copies of these sets through isomorphism, for example, let define $\mathbb C := \mathbf C$ then $$ \mathbb R := \{ x\in\mathbb C \:|\: \exists x'\in\mathbf R : x = \phi_{\mathbf R}(x')\} $$ with $\phi_{\mathbf R} : \mathbf R \to \mathbf C =:\mathbb C$ an isomorphism. This set $\mathbb R$ is a subset of $\mathbb C$. Moreover, it's a field and we can call it the "set of real numbers", and so on... ?

Then, if I want to construct quaternion $\mathbb H$, $\mathbb C$ become some copy of $\mathbf C$ and a subset of $\mathbb H$, et cætera...

Asaf Karagila
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Célestin
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  • I would advise against this notation in boldface for what you want to do: $\mathbf{N, Z, Q, R, C, H}$ is already used in old-style mathematics notation , in the place of $\mathbb{N,Z,Q,R,C,H}$… – Bernard Mar 11 '17 at 12:51
  • In this case, just denoting them by $N$, $Z$, $Q$, $R$, $C$ and $H$, it doesn't really matter... – Célestin Mar 11 '17 at 12:54
  • But what's the point of defining them a real subsets? Everything is defined within isomorphisms anyway. If you really want it, consider the images of all those guys in the last one, $\mathbf H$. – Bernard Mar 11 '17 at 12:57
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    It is not an answer, but a tackling details: Peano axioms does not provide a set of natural numbers, though it define what a natural number is. We need an actual construction of natural numbers from sets to provide natural numbers from set theory. – Hanul Jeon Mar 11 '17 at 12:58
  • Under ZFC, natural numbers are constructed with von Neumann ordinals (using the Axiom of Infinity), right ? – Célestin Mar 11 '17 at 13:07
  • Yes, you are right. – Hanul Jeon Mar 11 '17 at 13:42
  • Does it possible to define $\mathbb R$ as a subset of $\mathbb C$, then $\mathbb Q$ as a subset of $\mathbb R$ and so on... through composition of isomorphism (still an isomorphism) with my definition of $\mathbb C$, otherwise, if I want another field including $\mathbb C$, I can redefine this set to be a subset of the "greatest" (e.g. to be a subset of $\mathbb H$) ? – Célestin Mar 11 '17 at 13:51

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Most mathematician will be happy with just an embedding (i.e. injective homomorphism), not the exact inclusion relationship. However, we can make the set of reals really contains the set of natural numbers, by cutting copies of the image of embedding map $i:\mathbb{N}\to \mathbb{R}$ and putting the set of natural numbers.

Here is a detail: Let $\mathbb{R}_0$ be a set of real numbers we constructed (via Cauchy sequence or Dedekind cut, it doesn't matter.) with an embedding $i:\mathbb{N}\to\mathbb{R}_0$. Define $\mathbb{R} = (\mathbb{R}_0\setminus i[\mathbb{N}])\cup\mathbb{N}$ with appropriate operations and relations; for example we define the addition of $\mathbb{R}$ as

$$a+b := \begin{cases} i(a+b) &\text{if } a, b\in \mathbb{R}_0\text { and }a+b\notin i[\mathbb{N}] \\ i(a)+b& \text{if } a\in \mathbb{N} \text{ and } b\in \mathbb{R}_0 \\ a+i(b)& \text{if } a\in \mathbb{R}_0 \text{ and } b\in \mathbb{N} \\ a+b &\text{otherwise} \\\end{cases}$$ (Note that we abuse notations: the addition operations that appeared are not same, though they use same symbol.)

and so on. Of course, we should care that our $\mathbb{R}_0$ and $\mathbb{N}$ have empty intersection. If not our addition could be ill-defined.

Hanul Jeon
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  • Cutting seems interesting and funny, but showing that $\mathbb Z$ is a ring, $\mathbb Q$ and $\mathbb R$ are field must be longer... – Célestin Mar 11 '17 at 15:28
  • "Cutting" doesn't work because inverse operations are ill-defined... – Célestin Mar 13 '17 at 20:22
  • @pheonix No, we can define inverse operations similarly. Definimg it is just a tedious thing – Hanul Jeon Mar 14 '17 at 01:55
  • Well $+$ and $\times$ are ill-defined too, because cutting $i[\mathbb N]$ implies that $a+{\mathbb Z} 0 = i(a)+{\mathbb Z_0}0_{\mathbb Z_0}$ is not defined in $\mathbb Z$... – Célestin Mar 15 '17 at 11:26
  • @Phoenix Well, my definition might not work, however, the main idea is substitute elements of $i[\mathbb{N}]$ to corresponding ones of $\mathbb{N}$. Substituting elements could be tedious and dirty but it works. – Hanul Jeon Mar 15 '17 at 11:29
  • The problem isn't the definition of the operation... it's the definition of the set itself :/ – Célestin Mar 15 '17 at 11:32
  • @Phoenix I implicitly assume that $\mathbb{R}_0$ and $\mathbb{N}$ is disjoint. It might not hold, but it can be easily resolved; for example, we can consider $\mathbb{R}_0\times {0}$ instead of $\mathbb{R}_0$. Please let me know there is another problem. – Hanul Jeon Mar 15 '17 at 11:35