# Set Theoretic Definition of Numbers

I am reading the book by Goldrei on Classic Set Theory. My question is more of a clarification. It is on if we are overloading symbols in some cases. For instance, when we define $2$ as a natural number, we define When we define $2$ as an integer, $2_{\mathbb{Z}}$ is an equivalence class of ordered pair Similarly, when we define $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair and as a real number we define it as the left Dedekind cut of rationals less than $2_{\mathbb{Q}}$, i.e.

The clarification is each of the above are different objects right? So when we say $2$, it depends on the context? Also, if the above is true, is it correct or incorrect to say that "The set of natural numbers is a subset of reals"? Should we take the statement with a pinch of salt and understand accordingly?

Yes. And no.

You start with $\mathbb{N}$, and define $+$ and $\times$ (and $\lt$ and so on) appropriately.

Then you define an equivalence relation on $\mathbb{N}\times\mathbb{N}$ given by

and call the quotient set $(\mathbb{N}\times\mathbb{N})/\sim$ by the name $\mathbb{Z}$. (Behind the scenes, we are thinking of $(a,b)$ as meaning "the solution to $a=x+b$").

We can then define an addition $+_{\mathbb{Z}}$ and a product $\times_{\mathbb{Z}}$ on $\mathbb{Z}$, as well as an order $\leq_{\mathbb{Z}}$ by

and show that this is well defined. (I am using $[(a,b)]$ to denote the equivalence class of the pair $(a,b)$.

Certainly, $\mathbb{N}$ and $\mathbb{Z}$ are entirely different animals; set-theoretically, you can even show that they are disjoint.

But we can define a map $f\colon \mathbb{N}\to\mathbb{Z}$ by $f(n) = [(n,0)]$. This map is one-to-one, and for all $n,m\in\mathbb{N}$,

That means that even though $\mathbb{N}$ and $\mathbb{Z}$ are disjoint, there is a "perfect copy" of $\mathbb{N}$ (in so far as its operations $+$ and $\times$ are concerned, and as far as the order $\leq$ is concerned) sitting inside of $\mathbb{Z}$. (Added: In fact, this $f$ not only gives us perfect copy, it is the only map from $\mathbb{N}$ to $\mathbb{Z}$ that is one-to-one and respects all the operations; we say it is a "canonical embedding"). Since we have this perfect copy, and a very specific map identifying this copy with the original, we can think of $\mathbb{N}$ as being a subset of $\mathbb{Z}$ by "identifying it with its copy". So we do that. We can then introduce notation by showing that for every $[(a,b)]\in\mathbb{Z}$, either $a=b$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(a,b)]=f(n)=[(n,0)]$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(b,a)]=f(n)=[(n,0)]$; and then using $0$ to denote the class with $a=b$, $n$ to denote the class with $[(a,b)]=[(n,0)]$, and $-n$ to denote the class $[(c,d)]$ with $[(d,c)]=[(n,0)]$. This notation makes the identification clearer.

Similarly, once we have $\mathbb{Z}$, we define $\mathbb{Q}$ as the quotient of $\mathbb{Z}\times(\mathbb{Z}-\{0\}$ modulo $\cong$, where

(behind the scenes, we think of $(a,b)$ as meaning "the solution to $a=xb$").
We can then proceed as before, defining

and showing this is well defined; defining an order, etc. Again, $\mathbb{Q}$ and $\mathbb{Z}$ (and the original $\mathbb{N}$) are completely different sets. But we have a function $g\colon\mathbb{Z}\to\mathbb{Q}$ defined by $g(a) = [(a,1)]$. This is one-to-one, $g(a+_{\mathbb{Z}}b) = g(a)+_{\mathbb{Q}}g(b)$, and $g(a\times_{\mathbb{Z}}b) = g(a)\times_{\mathbb{Q}}g(b)$. (Added: And again, this is the only map from $\mathbb{Z}$ to $\mathbb{Q}$ that satisfies these conditions.) So again, we have a "perfect copy" of $\mathbb{Z}$ sitting inside of $\mathbb{Q}$ (and so also a perfect copy of the perfect copy of $\mathbb{N}$ that is sitting inside of $\mathbb{Z}$). So once again we "identify" $\mathbb{Z}$ with its image inside $\mathbb{Q}$ (and so we identify $\mathbb{N}$ with its image inside the image of $\mathbb{Z}$). Because, via $f$ and $g$, we have perfect copies of them anyway.

We do the same thing with $\mathbb{Q}$ as being "inside of $\mathbb{R}$", by identifying elements of $\mathbb{Q}$ with specific Dedekind cuts or with specific equivalence classes of Cauchy sequences, showing the identification is one-to-one and respects all the operations (and is essentially unique), and so obtaining a "perfect copy" of $\mathbb{Q}$ inside of $\mathbb{R}$ (and by extension, perfect copies of $\mathbb{N}$ and of $\mathbb{Z}$ also sitting inside of $\mathbb{R}$).

You can keep going, of course: define $\mathbb{C}$ as the set of all pairs $\mathbb{R}\times\mathbb{R}$; then identify $\mathbb{R}$ with the pairs $\mathbb{R}\times\{0\}$, and you have a copy of $\mathbb{N}$ sitting inside a copy of $\mathbb{Z}$ sitting inside a copy of $\mathbb{Q}$ sitting inside a copy of $\mathbb{R}$ sitting inside $\mathbb{C}$. (And then you can stick $\mathbb{C}$ inside the quaternions, the quaternions inside the octonions).

So even though they are actually very different sets, we have copies of each sitting inside the "next one", copies that respect all the structures we are interested in, so we can still think of them as being "subsets".

You used to do that all the time without the formalism: we think of "fractions" as being made up of an integer, a solidus, and a nonzero integer, so that "$3$" is not a fraction; but when needed, we are perfectly happy writing "$3 = \frac{3}{1}$" and working with either version of $3$ (the integer, or the fraction) depending on context.