Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \}$$ When we define $2$ as an integer, $2_{\mathbb{Z}}$ is an equivalence class of ordered pair $$2_{\mathbb{Z}} = \{(n+_{\mathbb{N}}2_{\mathbb{N}},n):n \in \mathbb{N}\}$$ Similarly, when we define $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair $$2_{\mathbb{Q}} = \{(a \times_{\mathbb{Z}} 2_{\mathbb{Z}},a):a \in \mathbb{Z}\backslash\{0\}\}$$ and as a real number we define it as the left Dedekind cut of rationals less than $2_{\mathbb{Q}}$, i.e. $$2_{\mathbb{R}} = \{q \in \mathbb{Q}: q <_{\mathbb{Q}} 2_{\mathbb{Q}}\}$$

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?

share|improve this question
3  
You might find this "buzz" by Terence Tao useful: google.com/buzz/114134834346472219368/RarPutThCJv/… –  Jon Dec 19 '10 at 4:24
5  
Your understanding is absolutely correct. The thing is that when one defines the integers, they come with a canonical embedding of $\mathbb{N}$. Similarly, the rationals come with a canonical embedding of $\mathbb{Z}$ and so on. So it does make sense to speak of subsets. Indeed, it becomes so cumbersome to distinguish between genuine subsets and images under an embedding that one fairly quickly drops this distinction in practice, unless one really has to thing about the foundations. –  Alex B. Dec 19 '10 at 4:34
    
Thanks Jon and Alex. –  user17762 Dec 19 '10 at 5:48
2  
This is a very good question. –  Asaf Karagila Dec 19 '10 at 7:37
    

3 Answers 3

up vote 22 down vote accepted

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 $$(a,b)\sim(c,d) \Longleftrightarrow a+d=b+c,$$ 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 $$\begin{array}{rcl} [(a,b)]+_{\mathbb{Z}}[(c,d)] &=& [(a+c,b+d)]\\\ [(c,d)]\times_{\mathbb{Z}}[(c,d)] &=& [(ac+bd,ad+bc)]\\\ [(a,b)] \leq [(c,d)] &\Leftrightarrow& a+d\leq b+c, \end{array}$$ 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 $n,m\in\mathbb{N}$, $$\begin{array}{rcl} f(n+m) &=& f(n)+_{\mathbb{Z}}f(m),\\\ f(n\times m) &=& f(n)\times_{\mathbb{Z}}f(m)\\\ n\leq m &\Leftrightarrow& f(n)\leq_{\mathbb{Z}} f(m) \end{array}$$ 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 $$(a,b)\cong (c,d) \Longleftrightarrow ad=bc$$ (behind the scenes, we think of $(a,b)$ as meaning "the solution to $a=xb$"). We can then proceed as before, defining $$\begin{array}{rcl} [(a,b)]+_{\mathbb{Q}}[(c,d)] &=& [(ad+bc,bd)]\\\ [(a,b)]\times_{\mathbb{Q}}[(c,d)] &=& [(ac,bd)] \end{array}$$ 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.

share|improve this answer
2  
It is important to emphasize not only the existence but also the uniqueness of the embedding of the subobject. Otherwise the overloading could be ambiguous. E.g. in rings w/o 1 does 2 denote (2,0) or (0,2) or (2,2) in Z x Z ? Note that this example also emphasizes another point: that the implicit "context" must include info such as the type of the structure(s) (here rings vs. rngs). Thus the overloaded notation is valid if there exists a canonical embedding as objects of the (implicitly) specified type. –  Bill Dubuque Dec 19 '10 at 15:03
    
@Arturo Magidin , your answer is great! I think it had answered a question which I always aked for myself ! thanx! –  Maths Lover Sep 20 '13 at 1:09

Yes, in the sense that in ZF, each of the above definitions defines a different collection of sets. Also yes, in the sense that one should always be aware of context when discussing a mathematical object.

The mathematical fact that allows us to overload the symbol $2$ here is the existence of a sequence of natural injections $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$, and elements in any term in this sequence can be pushed forward under the next injection or, if they lie in the range of the previous injection, pulled back. Most people do this automatically and so don't usually point it out explicitly. More generally, $1$ refers to the multiplicative identity in any ring, $0$ to the additive identity in any ring, and $2 = 1 + 1, 3 = 1 + 1 + 1$, and so forth.

share|improve this answer
    
Thanks –  user17762 Dec 19 '10 at 5:55

In addition to the other answers, it's also noteworthy to know that there is a number system (more-or-less) integrating N, Z, Q and R (and a lot more, but not C!).

The surreal numbers take the basic idea from Dedekind cuts, assigning to each number a so-called left set ('smaller') and right set ('larger') of numbers (constraint to certain rules), bootstrapping the whole process from the empty set and ending up with the reals and weird infinitesimals like $\frac{1}{\sqrt{\omega - \pi}}$!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.