# Convergence of monotone nets

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?

• Since I'm not sure what topology I had in mind when I wrote it, I changed it to the case of the linear order topology. – Keen-ameteur Jan 13 at 19:33

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 $$c, $$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)$$.
• Minor nitpick: $L$ could be $\max X$ and then its neighbourhoods are of the form $(c,L]$. Doesn't really affect the proof, the same argument can be used for both types. – Henno Brandsma Jan 13 at 23:09