# Real number (as Dedekind cut)

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


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 $$\sqrt[3]{2}$$), 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).