A closed bounded set contains its supremum I want to make sure that my reasoning is correct for this more simple proof. I am studying for a final exam in real analysis and don't want to take any chances so any feedback is appreciated.
Suppose $S \subseteq \mathbb{R}$, is a closed bounded set. Then $Sup(S)\in S$.
Proof.
If $S$ is empty, then $Sup(S)$ does not exist. So suppose $S$ is non-empty.
Then $\exists \alpha =Sup(S)$. Suppose $\alpha \notin S$. Let {$x_n$} be a sequence of points in $S$ such that $\alpha-\frac{1}{n}<x_n<\alpha$ for each natural number $n$. Then $\lim_{n\to\infty} x_n =\alpha \notin S$. But $S$ is closed so it contains all of its limit points, so we get a contradiction.
 A: The proof is correct. Two remarks:


*

*You shouldn't have written “$\exists\alpha=\sup S$”. It's much better to write: “then $\sup S$ exists; let us call it $\alpha$”.

*There is no need to prove it by contradiction. You could have eliminated the sentence “Suppose $\alpha\notin S$” and then end the proof by proving that $\alpha\in S$.

A: This is a duplicate of this other MathSE question, and one of the answers uses reasoning very similar to yours. Your proof is not completely correct, but it's close. There are a few things you ought to consider to make a proof with this idea:


*

*If $S$ is finite, the result is trivial (why is this the case?).

*When constructing your sequence, you should use non-strict inequalities, i.e. $\alpha - \frac{1}{n} \leq x_n \leq \alpha$. Why? If you don't, then you might not be able to choose an $x_n$ for every $n$. Consider the set $S = \{0,1\}$, with $\text{sup }S = 1$. There's no element of $S$ that you could assign to $x_1$ (or $x_2$, or any $x_n$) if you use strict inequalities in your definition. If you'd rather use an infinite set, take $S = \{\frac{1}{n}: n \in \mathbb{Z}_+\} \cup \{0\}$.

*This is more stylistic, but you should explain why there exists an $x_n$ that satisfies $\alpha - \frac{1}{n} \leq x_n \leq \alpha$ for every $n$. When you start to think about the justification for this, you'll avoid issues like using $<$ instead of $\leq$.

