In sequences of real numbers, we have a monotone convergence result:
If $a_{n+1}\geq a_n$ and bounded, then $a_n$ converges to it's supremum.
The proof seems to work also in the net case. My question is given that our net is not into the reals but a general linearly ordered space, and it is a monotonically increasing and bounded, can we say that such always converges in the order topology to it's supremum?
Yes. Let $(a_i)$ be an increasing net in a totally ordered set $X$, and suppose the supremum $L=\sup \{a_i\}$ exists. Then I claim $L$ is the limit of $(a_i)$.
Indeed, let $(c,d)$ be any open interval around $L$ (I include the possibility that $c=-\infty$ or $d=\infty$). Since $c<L$, $c$ is not an upper bound for $\{a_i\}$, so there exists some $i$ such that $a_i>c$. Then for any $j\geq i$, $L\geq a_j\geq a_i$, and in particular $a_j\in (c,d)$. That is, the net $(a_i)$ is eventually in the open interval $(c,d)$.