Prove if ${x_n}$ is monotone increasing and bounded, then $\lim_{n\to \infty} x_n$ = sup${x_n:n\in N}$.
Suppose there exists a $B$ such that $x_n\le B$ for all $n$, and let $x$ = sup${x_n: n\in N}$.
Let $\varepsilon \gt 0$ be arbitrary.
As $x$ is the supremum, then there must be at least one $M \in N$ such that $x_M \ge x-\varepsilon$ (because $x$ is the supremum) .
As ${x_n}$ is monotone increasing, then it is easy to see (by induction) that $x_n \ge x_M$ for all $n \ge M$.
Hence, $|x_n-x|=x-x_n\le x-x_M\lt \varepsilon$.
I don't understnad one part of this proof. If $x$ is larger than B, and $\varepsilon$ is really small, $x-\varepsilon$ might be still larger than B, but how do we know that there must be at least one $M$ such that $x_M \ge x-\varepsilon$ ??