The reals form a totally ordered field

Positivity is independent of the representative: if $a_n>\delta$ for $n\ge N$ and $(a_n^{\mathrm{\prime }}-a_n)$ is null, then beyond some $N^{\mathrm{\prime }}\ge N$ also $\mathrm{\mid }{a}_n^{\mathrm{\prime }}-a_n\mathrm{\mid }<\delta \mathrm{/}2$, so $a_n^{\mathrm{\prime }}>\delta \mathrm{/}2$: the defining property holds for $(a_n^{\mathrm{\prime }})$ with $\delta \mathrm{/}2$.

Trichotomy: if $x\mathrm{\ne }0$, any representative is non-null, so by the dichotomy either $a_n>\delta$ eventually ($x$ positive) or $a_n<-\delta$ eventually ($-x$ positive); the two exclude each other, and exactly one of $x$ positive, $x=0$, $-x$ positive holds.

Positives are closed under $+$ and $\cdot$: from $a_n>\delta$ and $b_n>{\delta }^{\mathrm{\prime }}$ eventually, $a_n+b_n>\delta +{\delta }^{\mathrm{\prime }}$ and $a_n{b}_n>\delta {\delta }^{\mathrm{\prime }}$ eventually, with $\delta +{\delta }^{\mathrm{\prime }},\delta {\delta }^{\mathrm{\prime }}>0$.

Consequently $\le$ is a total order (trichotomy plus transitivity from closure under sums), compatible with addition (translation preserves the difference) and with multiplication by positives: $\mathbb{R}$ is a totally ordered field.