Let F be an ordered field with least upper bound property.
1.Let $\alpha: \mathbb{N} \to F$ be a Cauchy sequence. Since F is an ordered field, $x$ is bounded both above and below.
2.By assumption and dual of it, $A$={$\alpha(n)$|$n\in \mathbb{N}$} has a inf $a_0$ and sup $b_0$.
3.F is Archimedean
4.If subsequence of a Cauchy sequence is convergent to $a\in F$ then the Cauchy sequence is convergent to $a\in F$
These are all i know.. How do I prove all Cauchy sequences are convergent in $F$?
Please consider my level. I want quite a direct proof not mentioning any topology & Cauchy net.
*Comment button is not available to me now, (I don't know why), so i write this here. I just proved it with facts that (i)every cauchy sequence is convergent in the set of Cauchy reals and (ii)there exists a bijective homomorphism between two dedekind complete fields and (iii)the set of Cauchy reals is dedekind complete. Let $x:i→x(i):\mathbb{N}→F$ be a cauchy sequence in dedekind complete field $F$. Then use the bijective homomorphism $f$ to show that $x':i→f(x(i))$ is a cauchy sequence in the set of Cauchy reals. By the fact (i), $x'$ is convergent. Since inverse of $f$ is also homomorphism, use this to show that $x$ is convergent.