Dedekind complete ⇒ Sequentially complete 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.

Let $c_0 = \frac{a_0 + b_0}2$ (Note that an ordered field has characteristic 0, so $2 = 1+1$ is invertible in $F$). If $[a_0, c_0]$ contains infinitely many $\alpha(n)$, then let $a_1 = a_0$, $b_1 = c_0$, otherwise $a_1 = c_0$, $b_1 = b_0$. Now set $c_1 = \frac{a_1 + b_1}2$. 
Continuing inductively, we obtain sequences $(a_n)$, $(b_n)$ such that: 


*

*$[a_n, b_n]$ contains infinitely many $\alpha(k)$.

*$b_n - a_n = \frac{b_0 - a_0}{2^n}$

*$a_{n} \le a_{n+1} \le b_{n+1} \le b_n$.


hold for each $n$.
Now let $a^* = \sup_n a_n$ (note that $a_n \le b_0$, so the sup exists) and $b_* = \inf b_n$. Then $a_n \le a^* \le b_* \le b_n$ for each $n$, so $b_* - a^* \le b_n - a_n = \frac{b_0 - a_0}{2^n}$. As $F$ is Archimedian, $b_* - a^* = 0$, i. e. $b_* = a^*$.
Set $k_0 := 1$, and for each $n$, given $k_n$ choose $k_{n+1} > k_n$ with $\alpha(k_{n+1}) \in [a_n, b_n]$ (which is possible since the latter interval contains infinitely many $\alpha(k)$). 
We will prove, that $\alpha(k_n) \to a_*$ (which, by 4., suffices). So let $\epsilon > 0$, as $F$ is Archimedian, there is some $N$ with $2^{-N} \le \epsilon(b_0 - a_0)$. For $n \ge N$ we now have $\alpha(k_n) \in [a_N, b_N]$ and $a^* \in [a_N, b_N]$, hence $|a_n - a^*| \le b_N - a_N \le 2^{-N}(b_0 - a_0) \le \epsilon$.
