There exists strictly increasing $\{x_n\}$ that converges to $\sup E$ I need to prove that, if $E \subseteq \mathbb{R}$ is a non-empty bounded set and $\sup E \not\in E$ then there exists a strictly increasing sequence $\{x_n\}$ that converges to $\sup E$ such that $x_n \in E$ for all $n \in \mathbb{N}$.
I've been trying to find a clue in the textbook, but couldn't. I don't even know how to start the proof. Could someone please give a clue?
 A: Hint 
Firstly, choose any element $x_1 \in E$. As stated, $\sup E \notin E \implies x_1 < \sup E$. Using the approximation property for suprema, for each $n \geq 2 \exists x_n \in E$ such that $\max(\sup E -\frac{1}{n},x_{n-1})<x_n<\sup E$ so that a sequence $x_1<x_2<x_3<...$ is strictly increasing.
Then go on to apply the Squeeze theorem and you should be home and dry.
Edit Show that the sequence above is increasing 
Start at $n=1$ there exists an $x \in E$ such that $\sup  E  − 1 < x$. Denote this $x = x_1$. Now, for $n=2$ there exists an $x \in E$ such that $\sup E − \frac{1}{2}  < x_2$. Clearly $x_2 > x_1$. This works for any $n$ so there exists a sequence $x_n \in E$ with the property that $\sup E - \frac{1}{n}<x_{n}$
A: Hint. $E\cap(\sup E-x_{n-1},\sup E]$ is not empty.
A: I'll turn my hint into an answer.
Start with a sequence $x_n$ in $E$ converging to $\sup E$. No element of that sequence can be greater than or equal to all the ones that follow, since $\sup E$ is not in $E$, so you can construct an increasing subsequence inductively: when you have chosen $L$ elements the last of which is some $x_k$, find an $x_n > x_k$ with $n > $k.
That increasing subsequence has the same limit as the original sequence, so you're done.
A: I would phrase a hint as this:
If there were no sequence approaching $\sup E$, how could $\sup E$ be the supremum?
