User:Abcxyz/Sandbox/Real Numbers/Ordering on Real Numbers is Total Ordering

Theorem
Let $\R$ denote the set of real numbers.

Let $\le$ denote the ordering on $\R$.

Then $\le$ is a total ordering.

Proof 1
Let $\left({\R, +, \times, \le}\right)$ denote the real numbers, as axiomatically defined as a Dedekind complete totally ordered field.

By definition, $\le$ is a total ordering.

Proof 2
Let $\R$ denote the set of real numbers, as constructed from Cauchy sequences.

Let $\le$ denote the ordering on $\R$.

Proof 3
Let $\R$ denote the set of real numbers, as constructed from Dedekind cuts.

We have that $\subseteq$ is an ordering (on $\R$).

It remains to show that $\subseteq$ is total (on $\R$).

Let $\alpha, \beta \in \R$.

Suppose that $\beta \nsubseteq \alpha$.

Then we can choose $p \in \beta$ such that $p \notin \alpha$.

Suppose that $q \in \alpha$.

Then $q < p$; otherwise, $p \in \alpha$.

Hence, $q \in \beta$.

That is, $\alpha \subseteq \beta$.

Proof 4
Let $\R$ denote the set of real numbers, as defined as the Dedekind completion of the rational numbers.

Let $\le$ denote the ordering on $\R$.

From Field of Rational Numbers, we have that $\left({\Q, \le}\right)$ is a totally ordered set.

The result follows from Dedekind Completion of Totally Ordered Set is Totally Ordered.