Let B be a bounded, nonempty subset of real numbers. Prove that there exists a sequence $X_n$ of real numbers such that for all $n\in{N},x_n\in{B}$ and $x_n\rightarrow\sup B$
My approach so far is that, because B is bounded there is an upperbound thus a supB and that it also converges. I also deducted that because Xn is a sequence of real numbers, the limit of Xn must also be defined as a real number as well. But I'm having trouble putting any actual math into this.
This is an analysis idea. Let $M = \sup(B) < \infty$, since $B$ is bounded.
By definition of the $\sup$ as the least upper bound, for any $\epsilon > 0$, we can find $b \in B$ such that $b > M - \epsilon$. Use this to construct the appropriate sequence $X_n$.