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 $\R$.

$\le$ is Reflexive
Let $x \in \R$.

It is to be shown that $x \le x$.

Let $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] = \left[{\!\left[{\left\langle{x'_n}\right\rangle}\right]\!}\right]$.

By Equivalence Class holds Equivalent Elements, we have $\left\langle{x_n}\right\rangle \sim \left\langle{x'_n}\right\rangle$.

By the definition of $\sim$, it follows that $x \le x$.

$\le$ is Transitive
Let $x, y, z \in \R$, $x \le y$, $y \le z$.

It is to be shown that $x \le z$.

Let $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right]$ and $z = \left[{\!\left[{\left\langle{z_n}\right\rangle}\right]\!}\right]$.

By definition of an equivalence class, we can choose $\left\langle{y_n}\right\rangle$ such that $y = \left[{\!\left[{\left\langle{y_n}\right\rangle}\right]\!}\right]$.

Let $\epsilon \in \Q_{>0}$.

Then, by the definition of $\le$, there exist $N_1, N_2 \in \N$ such that:
 * $\forall n \in \N: n > N_1 \implies x_n < y_n + \frac 1 2 \epsilon$


 * $\forall n \in \N: n > N_2 \implies y_n < z_n + \frac 1 2 \epsilon$

Let $N = \max {\left\{{N_1, N_2}\right\}} \in \N$.

Then:
 * $\forall n \in \N: n > N \implies x_n < z_n + \epsilon$

Hence, $x \le z$.

$\le$ is Antisymmetric
Let $x, y \in \R$, $x \le y$, $y \le x$.

It is to be shown that $x = y$.

By the definition of an equivalence class, we can choose $\left\langle{x_n}\right\rangle$ and $\left\langle{y_n}\right\rangle$ such that $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right]$ and $y = \left[{\!\left[{\left\langle{y_n}\right\rangle}\right]\!}\right]$.

By Equivalence Class holds Equivalent Elements, it suffices to show that $\left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle$.

Let $\epsilon \in \Q_{>0}$.

Then there exist $N_1, N_2 \in \N$ such that:
 * $\forall n \in \N: n > N_1 \implies x_n < y_n + \epsilon$


 * $\forall n \in \N: n > N_2 \implies y_n < x_n + \epsilon$

Let $N = \max {\left\{{N_1, N_2}\right\}} \in \N$.

Then:
 * $\forall n \in \N: n > N \implies -\epsilon < x_n - y_n < \epsilon$

It follows that $\left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle$.

$\le$ is Total
Let $x, y \in \R$.

It is to be shown that either $x \le y$ or $y \le x$.

Suppose that $y \not \le x$.

We have that $\left({\Q, \le}\right)$ is a totally ordered set.

It follows that there exist $\left\langle{\hat x_n}\right\rangle$, $\left\langle{\hat y_n}\right\rangle$, and $\hat \epsilon \in \Q_{>0}$ such that $x = \left[{\!\!\!\:\left[{\left\langle{\hat x_n}\right\rangle}\right]\!\!\!\:}\right]$, $y = \left[{\!\!\!\:\left[{\left\langle{\hat y_n}\right\rangle}\right]\!\!\!\:}\right]$, and:
 * $\forall M \in \N: \exists N \in \N: N > M$ and $\hat y_N \ge \hat x_N + \hat \epsilon$

Since $\left\langle{\hat x_n}\right\rangle$ and $\left\langle{\hat y_n}\right\rangle$ are Cauchy sequences, there exist $M_1, M_2 \in \N$ such that:
 * $\forall m, n \in \N: m, n > M_1 \implies \left\vert{\hat x_m - \hat x_n}\right\vert < \frac 1 4 \hat \epsilon$


 * $\forall m, n \in \N: m, n > M_2 \implies \left\vert{\hat y_m - \hat y_n}\right\vert < \frac 1 4 \hat \epsilon$

Let $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right]$, $y = \left[{\!\left[{\left\langle{y_n}\right\rangle}\right]\!}\right]$.

By Equivalence Class holds Equivalent Elements, it follows that there exist $N_1, N_2 \in \N$ such that:
 * $\forall n \in \N: n > N_1 \implies \left\vert{x_n - \hat x_n}\right\vert < \frac 1 4 \hat \epsilon$


 * $\forall n \in \N: n > N_2 \implies \left\vert{y_n - \hat y_n}\right\vert < \frac 1 4 \hat \epsilon$

Let $M = \max {\left\{{M_1, M_2, N_1, N_2}\right\}} \in \N$.

Then there exists $N \in \N$ such that $N > M$ and $\hat y_N \ge \hat x_N + \hat \epsilon$.

Let $n \in \N$, $n > N$.

It follows that:

Hence, $x \le y$.

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 $\R$.

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.