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?

  • $\begingroup$ 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. $\endgroup$ – 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 $c<L$, $c$ 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 $(c,d)$.

  • $\begingroup$ Just one further question. This also implies convergence when considering the left order topology? en.wikipedia.org/wiki/… $\endgroup$ – Keen-ameteur Jan 13 at 19:59
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    $\begingroup$ The left and right order topologies are both coarser than the order topology, so convergence with respect to the order topology implies convergence with respect to the left and right order topologies. $\endgroup$ – Eric Wofsey Jan 13 at 20:02
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Jan 13 at 23:09

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