In set theory, if natural numbers are represented by nested sets that include the empty set, how are the rest of the real numbers represented as sets?
Thanks for the answers. Several answers basically said for irrational numbers that
A Dedekind cut is a pair of sets of rational numbers $\{L, R\}$. The set of real numbers is defined to be the set of all Dedekind cuts, where a Dedekind cut is a pair of sets of rational numbers $\{L, R\}$ which have no elements in common, and where all the elements of $L$ are less than any element of $R$. Each Dedekind cut is a real number. This is where I have a problem – surely that can’t be correct. The set $L$ is a set of all rationals, and there must be a rational in the set $L$ that is greater than all other rationals in that set, even if we have no method of determining it. And similarly, there must be a rational in the set $R$ that is less than all other rationals in that set, even if we have no method of determining it. If every irrational number has a corresponding set $L$, then each irrational number has some such corresponding largest element of that set $L$, and then each irrational number has some corresponding rational number. And that would mean that the irrational numbers are countable. So, with Dedekind cuts, the only conclusion is that there must be irrational numbers $x$ which are either greater or lesser than some irrational cut $y$ of the rationals, and between $x$ and $y$ there is no rational number. But that is impossible, so that the Dedekind cuts cannot be the correct representation of the real numbers.
Surely the problem with Dedekind cuts is in using sets of rationals that include all rationals up to a certain rational. But there is an alternative method of representing irrationals can be defined in terms of infinite sets of rational numbers. For example, in binary notation, the non-integer part of $\pi$ is $.00100100\ 00111111\ 01101010\ 10001$. You define a set by: if the nth digit is a $1$, then the natural number $n$ is in the set. And then we have that, for the real numbers between $0$ and $1$, that the set of real numbers is simply the set of all subsets of natural numbers. Each subset corresponds to some real number between $0$ and $1$.
And in this way, all real numbers can be considered to be some set based only on nested sets of the empty set.
But I still haven’t got a satisfactory answer for how negative numbers can be represented in terms only of sets containing the empty set. Any ideas?
Best Answer
There are a few possibilities, but here is the one approach. Even the starting point—the set of natural numbers $\mathbb{N}$—can be defined in several ways, but the standard definition takes $\mathbb{N}$ to be the set of finite von Neumann ordinals. Let us assume that we do have a set $\mathbb{N}$, a constant $0$, a unary operation $s$, and binary operations $+$ and $\cdot$ satisfying the axioms of second-order Peano arithmetic.
First, we need to construct the set of integers $\mathbb{Z}$. This we can do canonically as follows: we define $\mathbb{Z}$ to be the quotient of $\mathbb{N} \times \mathbb{N}$ by the equivalence relation $$\langle a, b \rangle \sim \langle c, d \rangle \text{ if and only if } a + d = b + c$$ The intended interpretation is that the equivalence class of $\langle a, b \rangle$ represents the integer $a - b$. Arithmetic operations can be defined on $\mathbb{Z}$ in the obvious fashion: $$\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$$ $$\langle a, b \rangle \cdot \langle c, d \rangle = \langle a c + b d, a d + b c \rangle$$ (Check that these respect the equivalence relation.) Again, this is not the only way to construct $\mathbb{Z}$; we can give a second-order axiomatisation of the integers which is categorical (i.e. any two models are isomorphic). For example, we may replace the set $\mathbb{Z}$ by $\mathbb{N}$, since the two sets are in bijection; the only thing we have to be careful about is to distinguish between the arithmetic operations for $\mathbb{Z}$ and for $\mathbb{N}$. (In other words, $\mathbb{Z}$ is more than just the set of its elements; it is also equipped with operations making it into a ring.)
Next, we need to construct the set of rational numbers $\mathbb{Q}$. This we may do using equivalence relations as well: we can define $\mathbb{Q}$ to be the quotient of $\mathbb{Z} \times (\mathbb{Z} \setminus \{ 0 \})$ by the equivalence relation $$\langle a, b \rangle \sim \langle c, d \rangle \text{ if and only if } a d = b c$$ The intended interpretation is that the equivalence class of $\langle a, b \rangle$ represents the fraction $a / b$. Arithmetic operations are defined by $$\langle a, b \rangle + \langle c, d \rangle = \langle a d + b c, b d \rangle$$ $$\langle a, b \rangle \cdot \langle c, d \rangle = \langle a c, b d \rangle$$ And as before, we can give an axiomatisation of the rational numbers which is categorical.
Now we can construct the set of real numbers $\mathbb{R}$. I describe the construction of Dedekind cuts, which is probably the simplest. A Dedekind cut is a pair of sets of rational numbers $\langle L, R \rangle$, satisfying the following axioms:
The intended interpretation is that $\langle L, R \rangle$ is the real number $z$ such that $L = \{ x \in \mathbb{Q} : x < z \}$ and $R = \{ y \in \mathbb{Q} : z < y \}$. The set of real numbers is defined to be the set of all Dedekind cuts. (No quotients by equivalence relations!) Arithmetic operations are defined as follows:
John Conway gives an alternative approach generalising the Dedekind cuts described above in his book On Numbers and Games. This eventually yields Conway's surreal numbers.