Real number (as Dedekind cut)

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:

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 $S$ 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:

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.

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 $Q^{\le }$ and $Q^{\ge }$.

Completion of a space

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

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

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

1. Every rational number has a cube either greater than 2 or less than 2,
2. Because $f\left(x\right)=x^3$ is a monotonically increasing function, we have that $p<q⟺p^3<q^3$, which means that every element in $Q^{\le }$ is less than every element in $Q^{\ge }$.

So $(Q^{\le },Q^{\ge })$ is a Dedekind cut. However, there is no rational number whose cube is equal to $2$, so $Q^{\le }$ has no greatest element and $Q^{\ge }$ 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 $\sqrt[3]{32}$), which is "between" any two numbers in $Q^{\le }$ and $Q^{\ge }$.

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

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 $Q^{\ge }$. (If the cuts weren't one-sided, $r$ would also be mapped to the set where $r$ was the greatest element of $Q^{\le }$, 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 $Q^{\ge }$ are (strictly) greater than $q$.

Now we can define the comparison_operator $\le$ for two real numbers numbers $a=(Q_a^{<},Q_a^{\ge })$ and $b=(Q_b^{<},Q_b^{\ge })$ as follows: $a\le b$ 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.