# Showing a Sequence Converges to Sup $F$

In $\mathbb{R}$ with the usual metric, it holds that, for any $F \subset \mathbb{R}$ non-empty and bounded above, sup $F$ $\in \overline{F}$.

Show that there exists a sequence $S := (x_n)_{n \in \mathbb{N}}$ in $F$ with $x_n \rightarrow$ sup $F$ w.r.t. the usual metric.

I'd like to know if I'm approaching this problem correctly.

By definition, $\overline{F}$ = $F \bigcup F'$. Now, if sup $F \in \overline{F}$, then sup $F \in F$, sup $F \in F'$, or both.

If sup $F \in F'$, then by definition of $F'$ sup $F$ is an accumulation point. So, there must be some sequence that approaches sup $F$.

If sup $F \in F$, then this means sup $F$ is the maximum of $F$. So, $\forall f \in F$, $f \le$ sup $F$. Then we can take all terms of $F$ and construct that as a sequence. Since all the terms of this sequence are $\le$ sup $F$, the sequence approaches sup $F$.

Is this the correct way to approach this problem? I think I showed the sup $F \in F'$ part right, but I'm not very sure if the other case is true. I just tried to use the fact that sup $F \in F \implies$ sup $F$ is the maximum to try to "construct" a sequence with all the terms of $F$. Is this a valid approach? Or is there a simpler way to prove the problem. Thank you for your help.

I think it is more useful to use a more general approach, if you already know that $\sup E \in \overline{E}$.
It suffices to show that for every $x \in \overline{E}$, there exists a sequence $(x_n)_n$ such that $x_n \to x$.
By definition, $\forall \epsilon > 0: (x- \epsilon , x + \epsilon) \cap E \neq \emptyset$, so letting $\epsilon:= 1/n (n = 1,2, \dots)$, we can find a sequence $(x_n)_n$ in $E$ with $x_n \in (x-1/n, x+1/n)$ for all $n \geq 1$ meaning that $|x_n - x| < 1/n$ for all $n\geq 1$, and hence $x_n \to x$
• You are welcome! Note that the axiom of choice was used when picking elements out of every set in the collection $\{(x-1/n,x+1/n) \mid n \geq 1\}$ – user370967 Mar 4 '18 at 12:09