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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 the rational numbers are a dense space — between any two rational numbers you can always find infinitely many other rational numbers. How could there be gaps in a set like that?

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:

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

Every element of $S$ in $A$ is less than every rational number in $B$ .

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 two properties of a Dedekind cut:

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

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

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 $Q^{\leq}$ and $Q^{\geq}$ .

Completion of a space

If a space is complete (doesn't have any gaps in it), then in any Dedekind cut $(Q^{\leq}, Q^{\geq})$ , either $Q^{\leq}$ will have a greatest element or $Q^{\geq}$ will have a least element. (We can't have both at the same time — why?)

Show solution

Suppose $Q^{\leq}$ had a greatest element $q_{u}$ and $Q^{\geq}$ 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 $Q^{\leq}$ or $Q^{\geq}$ , contradicting property 1 of a Dedekind cut.

But in the rational numbers, we can find a Dedekind cut where neither $Q^{\leq}$ nor $Q^{\geq}$ have a greatest or least element respectively.

Consider the pair of sets $(Q^{\leq}, Q^{\geq})$ where $Q^{\leq} = {x \in Q | x^{3} \leq 2}$ and $Q^{\geq} = {x \in Q | x^{3} \geq 2}$ .

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

Because $f (x) = x^{3}$ is a monotonically increasing function, we have that $p < q ⟺ p^{3} < q^{3}$ , which means that every element in $Q^{\leq}$ is less than every element in $Q^{\geq}$ .

So $(Q^{\leq}, Q^{\geq})$ is a Dedekind cut. However, there is no rational number whose cube is equal to $2$ , so $Q^{\leq}$ has no greatest element and $Q^{\geq}$ 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 $\sqrt[3]{2}$ ), which is "between" any two numbers in $Q^{\leq}$ and $Q^{\geq}$ .

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 $(Q^{\leq}, Q^{\geq})$ with the additional property that the set $Q^{\leq}$ has no greatest element (in which case we now call it $Q^{<}$ ). The case where $Q^{\leq}$ has a greatest element $q_{g}$ can be trivially transformed into the equivalent case on the other side by moving $q_{g}$ to $Q^{\geq}$ where it is automatically the least element due to being less than any other element in $Q^{\geq}$ .

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 $Q^{\geq}$ . (If the cuts weren't one-sided, $r$ would also be mapped to the set where $r$ was the greatest element of $Q^{\leq}$ , which would make the mapping non-unique.)

An irrational number $q$ is newly defined by the Dedekind cut where all the elements of $Q^{<}$ are less than $q$ and all the elements of $Q^{\geq}$ are (strictly) greater than $q$ .

Now we can define the comparison_operator $\leq$ for two real numbers numbers $a = (Q_{a}^{<}, Q_{a}^{\geq})$ and $b = (Q_{b}^{<}, Q_{b}^{\geq})$ as follows: $a \leq b$ when $Q_{a}^{<} \subseteq Q_{b}^{<}$ .

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.

Proof

If $a = b$ , then $a \leq b$ and $b \leq a$ . By the definition of the comparison operator, we have that $Q_{a}^{<} \subseteq Q_{b}^{<}$ and $Q_{b}^{<} \subseteq Q_{a}^{<}$ , which means that $Q_{a}^{<} = Q_{b}^{<}$ , which means that the Dedekind cuts corresponding to $a$ and $b$ are also equal.

Proof of the field structure of Dedekind cuts.