Monotone convergence theorem for sequences in a completely metrizable totally ordered topological space In the Reals it is commonly known that a monotone bounded sequence is convergent. This is due to the first countability axiom of the metric space and the least upper bound property of the reals. Can we affirm that this is true also for sequences in a completely metrizable topological space, where the topology is the order topology generates by the total order?

No : take $\mathbb R \times \{0,1\}$ ordered lexicographically ($0<1$) with the following distance : $d((x,i), (y,j))=d(x,y)+ |i-j|$.
I claim that this metrizes the order topology, and is complete. Once we prove this, the claim will easily follow, because any Cauchy sequence must have, from some rank, the second coordinate constant, and then we use the fact that $\mathbb R$ is complete, and then that $(n,0)$ is monotone and bounded (by, say $(0,1)$) but does not converge.
To prove the claim, we show that opens in both topologies are open in the other one.
Let $U$ be open in the order topology, and $(x,i)\in U$. Then there are $a,b$ such that $(x,i)\in ]a,b[ \subset U$. Up to changing their second coordinates, we may assume $a,b$ both have $i$ as a second coordinate, and then we may take them small enough so that $a=(x-\epsilon, i), b = (x+\epsilon, i)$ with $\epsilon <1$.
Then $]a,b[ = B((x,i), \epsilon)$ which is open in the metric topology. So $U$ is open in the metric topology.
Conversely let $U$ be open in the metric topology and let $x\in U$. Then there is $\epsilon$ with $B(x, \epsilon)\subset U$. Up to taking $\epsilon <1$, $B(x,\epsilon)$ only consists of people with the same second coordinate as $x$, and so can be seen to be of the form $I\times \{i\}$, with $i$ the second coordinate of $x$, and $I$ an open interval of $\mathbb R$; which is open for the order topology. So $U$ is open for the order topology and we are done.
