I want to prove that any monotonically increasing sequence $(x_n)$ of extended real numbers converges to $\sup_{n} x_n$ (with respect to the order topology) .
There are two cases, either $(x_n)$ is bounded by some (positive) real number $M$ or not.
It's okay for the first case, using least-upper-bound property. But I stuck on the second one.
When the sequence $(x_n)$ is unbounded by a real, it means that for any positive real number $M$, there is some integer $n$ such that $x_n > M$. I want to show that $x_n \to \infty$.
Let $V_\infty$ be a neighborhood of $\infty$. I stuck a bit at this step, I want to show that there exists a positive real $M$ for which $]M, \infty]\subseteq V_\infty$. Since $(x_n)$ is increasing and unbounded, there will be some $n_0$ such that for any $n\geqslant n_0$, we have $x_n\geqslant x_{n_0}>M$, then $x_n \in V_\infty$ and we are done.
How to get the existence of at least one $M$ such that $]M,\infty] \subseteq V_\infty$ ? Thanks in advance.
What you have to prove is that for any open neighborhood $O$ of $\infty$, there is $N$ such that for all $n\ge N$, $x_n \in O$. If $O$ is an open set containing $\infty$, then by definition (of a basis of a topology), there is some $b\in \mathbb R$ such that $(b,\infty] \subset O$. Conclude.