Proof

We first show that the order relation for real numbers "$ < `$" is a trichotomy.

Let $x,y\in \mathbb R$ be any real numbers represented by the equivalence classes of some rational Cauchy sequences $(x_n)_{n\in\mathbb N}+ I$ and $(y_n)_{n\in\mathbb N}+ I$.
It follows from the definition of the order relation for real numbers that
- $x < y$ is equivalent to $x-y < 0$, which is equivalent to the existence of some $N\in\mathbb N$ such that $x_n-y_n < 0$ for all $n > N$.
- $x > y$ is equivalent to $x-y > 0$, which is equivalent to the existence of some $N\in\mathbb N$ such that $x_n-y_n > 0$ for all $n > N$.
- $x = y$ is equivalent to $x-y = 0$, which is equivalent to $(x_n-y_n)_{n\in\mathbb N}$ being a rational Cauchy sequence converging to $0$.
Please note that in the terms $x_n-y_n < 0$ and $x_n-y_n > 0$ we have reduced the order relations of real numbers "$x < y$" and "$x > y$" to the order relations of rational numbers "$x_n < y_n$" and "$x_n > y_n$", respectively.
Therefore, from the trichotomy of the order relation for rational numbers, the trichotomy for the order relation for real numbers follows.
Therefore, for any two given real numbers $x,y$, only one of the following cases can occure at once:
- Either $x=y$, or
- $x > y$, or
- $x < y.$

Now, we show the transitivity.

Let $x < y$ and $y < z$.
It follows $x - y < 0$ and $y - z < 0$.
Adding both sides of the inequations in the set of real numbers results in $x-y+y-z< 0$.
The existence of inverse real numbers with respect to addition results in $x- z < 0$.
This means that $x < y$.
The transitivity of "$>$", "$\ge$", and "$\le$" follows analogously.
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