I have a hard time proving the following statement:
Let $A$ be a non-empty bounded subset of $\mathbb R.$ Then there exists a convergent sequence $(x_n)_n$ with $x_n \in A$ for all $ \mathbf n \boldsymbol\in \mathbb N$ with its limit equal to $\boldsymbol{\sup}{A}$.
I have already proven that: $$\limsup_{n \to \infty }{x_n} \leq \sup{A}$$ for all sequences $(x_n)_n$ with it's elements in $A$. I tried assuming that all the sets in $A$ are divergent but I cannot seem to find a contradiction. I also know that each sequence of $A$ has a subsequence that is convergent with its limit being equal to $\limsup{x_n} = \liminf{x_n}$ because of the Bolzano–Weierstrass theorem, but there doesn't seem to be a good follow up for this to prove the statement. I don't need a full proof but any hints that can get me started are very much appreciated.
Thanks in advance !
We have $A\subset (-\infty, \sup A]$ because $\sup A$ is an upper bound for $A$.
If there existed $n\in \Bbb N$ such that $(-1/n+\sup A,\sup A]\cap A=\emptyset$ then (for such an $n$) we would have $A\subset (-\infty, -1/n+\sup A],$ implying that $A$ has an upper bound $-1/n+\sup A$ that's less than its $least$ upper bound $\sup A,$ which is absurd.
So to avoid absurdity, for each $n\in \Bbb N$ there must exist some $x_n\in (-1/n+\sup A,\sup A]\cap A.$