# Proof that the order axioms for Cauchy sequences are well defined

So the order axioms are the following:

For all $a, b, c \in S$ $$(O1) a < b \text{ or } a = b \text{ or } a>b$$ $$(O2) a<b \text{ and } b<c \therefore a<c$$ $$(O3) a < b \to a+c<b+c$$ $$(O4) a<b \text{ and } c>0 \therefore ac < bc$$

Now, since all Cauchy sequences are convergent sequences, let $[a_k], [b_k], [c_k]$ be the equivalence classes of the Cauchy sequences that converge to $a,b,c$ respectively. I mean, since $a,b,c \in \mathbb{R}$ and the reals are an ordered field, the order axioms follow automatically. But is this enough?

EDIT: The definition I am using is $a<b$ if $a_k<b_k$ eventually in a Cauchy sequence and viceversa with $a>b$.

• Wouldn't citing "the reals are an ordered field" be circular? – DHMO Apr 17 '17 at 7:43
• True. So, do I have to use $a',b',c'$ each also from their respective equivalence classes? For example for trichotomy, $a<b$ or $a=b$ or $a>b$ as well as $a'<b'$ or $a'=b'$ or $a'>b'$? – The Bosco Apr 17 '17 at 8:08
• Before trying to prove them, first define what it means for a sequence to be less than, equal to, or greater than another. And you can't use properties of real number for these definitions if the goal is to define real numbers using cauchy sequences. – genepeer Apr 17 '17 at 8:08
• @genepeer Oh sorry, I forgot to add that. The definition the book asks us to use is: $a<b$ if $a_k < b_k$ eventually in the Cauchy sequence and viceversa for $a>b$. – The Bosco Apr 17 '17 at 8:10
• It should be slightly different from that. That would mean (1, .1, .01, .001, ...) > (.1, .01, .001, .0001, ...) but you want them to actually be in the same equivalence class. Anyway, I'd say you shouldn't be trying to prove the axioms since they're axioms. Just take them as is and then use them to prove things about the real numbers, the set of equivalence classes. – genepeer Apr 17 '17 at 8:17