# 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:

• Real number
• Mathematics

Math­e­mat­ics is the study of num­bers and other ideal ob­jects that can be de­scribed by ax­ioms.

• 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).