# Real number (as Dedekind cut)

comment: Mnemonics for defined macros: \Ql = Q left, \Qr = Q right, \Qls = Q left strict, \Qrs = Q right strict.

The rational numbers have a problem that makes them unsuitable for use in calculus — they have “gaps” in them. This may not be obvious or even make sense at first, because between any two rational numbers you can always find infinitely many other rational numbers. How could there be gaps in a set like that? $$\newcommand{\rats}{\mathbb{Q}} \newcommand{\Ql}{\rats^\le} \newcommand{\Qr}{\rats^\ge} \newcommand{\Qls}{\rats^<} \newcommand{\Qrs}{\rats^>}$$ $$\newcommand{\set}{\left\{#1\right\}} \newcommand{\sothat}{\ |\ }$$

But using the construction of Dedekind cuts, we can suss out these gaps into plain view. A Dedekind cut of a totally ordered set $$S$$ is a pair of sets $$(A, B)$$ such that:

1. Every element of $$S$$ is in exactly one of $$A$$ or $$B$$. (That is, $$(A, B)$$ is a partition of $$S$$.)

2. Every element of $$A$$ is less than every element of $$B$$.

3. Neither $$A$$ nor $$B$$ is empty. (We’ll see why this restriction matters in a moment.)

One example of such a cut might be the set where $$A$$ is the negative rational numbers and $$B$$ is the nonnegative rational numbers (positive or zero). We see that it satisfies the three properties of a Dedekind cut:

1. Every rational number is either negative or nonnegative, but not both.

2. Every rational number which is negative is less than a rational number that is nonnegative.

3. There exists at least one negative rational number (e.g. $$-1$$) and one nonnegative rational number (e.g. $$1$$).

In fact, Dedekind cuts are intended to represent sets of rational numbers that are less than or greater than a specific real number (once we’ve defined them). To represent this, let’s call them $$\Ql$$ and $$\Qr$$.

Knowing this, why does it matter that neither set in a Dedekind cut is empty?

If $$\Ql$$ were empty, then we’d have a real number less than all the rational numbers, which is $$-\infty$$, which we don’t want to define as a real number. Similarly, if $$\Qr$$ were empty, then we’d get $$+\infty$$.

## Completion of a space

If a space is complete (doesn’t have any gaps in it), then in any Dedekind cut $$(\Ql, \Qr)$$, either $$\Ql$$ will have a greatest element or $$\Qr$$ will have a least element. (We can’t have both at the same time — why?)

Suppose $$\Ql$$ had a greatest element $$q_u$$ and $$\Qr$$ had a least element $$q_v$$. We can’t have $$q_u = q_v$$, because the same number would be in both sets. So then because the rational numbers are a dense space, there must exist a rational number $$r$$ so that $$q_u < r < q_v$$. Then $$r$$ is not in either $$\Ql$$ or $$\Qr$$, contradicting property 1 of a Dedekind cut.

But in the rational numbers, we can find a Dedekind cut where neither $$\Ql$$ nor $$\Qr$$ have a greatest or least element respectively.

Consider the pair of sets $$(\Ql, \Qr)$$ where $$\Ql = \set{x \in \rats \mid x^3 \le 2}$$ and $$\Qr = \set{x \in \rats \mid x^3 \ge 2}$$.

1. Every rational number has a cube either greater than 2 or less than 2,

2. Because $$f(x) = x^3$$ is a monotonically increasing function, we have that $$p < q \iff p^3 < q^3$$, which means that every element in $$\Ql$$ is less than every element in $$\Qr$$.

So $$(\Ql, \Qr)$$ is a Dedekind cut. However, there is no rational number whose cube is equal to $$2$$, so $$\Ql$$ has no greatest element and $$\Qr$$ has no least element.

This represents a gap in the numbers, because we can invent a new number to place in that gap (in this case $$\sqrt{2}$$), which is “between” any two numbers in $$\Ql$$ and $$\Qr$$.

## Definition of real numbers

Before we move on, we will define one more structure that makes the construction more elegant. Define a one-sided Dedekind cut as any Dedekind cut $$(\Ql, \Qr)$$ with the additional property that the set $$\Ql$$ has no greatest element (in which case we now call it $$\Qls$$). The case where $$\Ql$$ has a greatest element $$q_g$$ can be trivially transformed into the equivalent case on the other side by moving $$q_g$$ to $$\Qr$$ where it is automatically the least element due to being less than any other element in $$\Qr$$.

Then we define the real numbers as the set of one-sided Dedekind cuts of the rational numbers.

• A rational number $$r$$ is mapped to itself by the Dedekind cut where $$r$$ itself is the least element of $$\Qr$$. (If the cuts weren’t one-sided, $$r$$ would also be mapped to the set where $$r$$ was the greatest element of $$\Ql$$, which would make the mapping non-unique.)

• An irrational number $$q$$ is newly defined by the Dedekind cut where all the elements of $$\Qls$$ are less than $$q$$ and all the elements of $$\Qr$$ are (strictly) greater than $$q$$.

Now we can define the total order $$\le$$ for two real numbers $$a = (\Qls_a, \Qr_a)$$ and $$b = (\Qls_b, \Qr_b)$$ as follows: $$a \le b$$ when $$\Qls_a \subseteq \Qls_b$$.

Using this, we can show that unlike in the Cauchy sequence definition, we don’t need to define any equivalence classes — every real number is uniquely defined by a one-sided Dedekind cut.

If $$a = b$$, then $$a \le b$$ and $$b \le a$$. By the definition of the order, we have that $$\Qls_a \subseteq \Qls_b$$ and $$\Qls_b \subseteq \Qls_a$$, which means that $$\Qls_a = \Qls_b$$, which means that the Dedekind cuts corresponding to $$a$$ and $$b$$ are also equal.

Proof of the field structure of Dedekind cuts.

Children:

Parents:

• I think that every metric space is dense in itself. If X is a metric space, then a set E is dense in X whenever every element of X is either a limit point of E or an element of E (or both).