TITLE: Question related with Archimedean property and complete ordered field.
QUESTION [0 upvotes]: Let $K$ be an ordered field, in which the Archimedean property is satisfied.
Show that if every bounded increasing sequence in $K$ has a least
upper bound in $K$, then $K$ is a complete ordered field.

REPLY [0 votes]: I agree, the process your professor mentions doesn't have to terminate.
Example: Take $(a_{1i})_i$ as an increasing sequence converging to $a_1 = 1$, $(a_{2i})_i$ converge to $a_2 = a_1 + \frac{1}{2^2}$ and inductively $(a_{ji})_i$ a sequence converging to $a_j = a_{j-1} + \frac{1}{j^2}$. This is a sequence you could obtain in the process for the bounded set $(0,10) \subseteq \mathbb{R}$ and obviously does not terminate. By some transfinite induction argument this might get fixed (someone else has to verify this).
How about you define a sequence as follows: Take $a_0 \in A$ arbitrary.
If $a_0 + 1$ is not an upper bound for $A$, then there is some $a_0 + 1 < a_1 \in A$. We continue this process and by the archemedian property we obtain some $a_n$ such that $a_n + 1$ is an upper bound for $A$. Denote this by $b_1 = a_n$.
Now repeat the process with $a_0 = b_1$ and instead of adding $1$ we add $\frac{1}{2}$ and later $\frac{1}{3}, \frac{1}{4}\ldots$
This yields an increasing sequence $(b_i)_i$ with least upper bound $b$. We show that $b$ is also a least upper bound for $A$. Assume there is $a \in A$ with $a > b$. By the archemedian principle there is $n$ with $a - b > \frac{1}{n}$ or, in other words, $a > b + \frac{1}{n} > b_n + \frac{1}{n} \geq a$ by definition of $b_n$. Thus $b$ is in fact an upper bound for $A$. Now assume there is an upper bound $b'$ of $A$ with $b' < b$. Again by the archemedian principle we have $n$ such that $b' + \frac{1}{n} < b \leq b_n + \frac{1}{n}$ and thus $b' < b_n \in A$ which is again a contradiction.
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