General topology / real analysis - Suppose $S$ is a bounded and closed nonempty subset of real numbers. Prove sup S is in S

Question

Q. Suppose S is a bounded and closed nonempty subset of real numbers. Prove $\sup S$ is in $S$.

Since $S$ is bounded and by the least upper bound property of $\mathbb R$, there exists $\sup S \in \mathbb R$ which we do not know if it is in $S$ yet.

By the definition of supremum, for $\epsilon \gt 0$,

$\exists s \in S$ such that, $\sup S - \epsilon \lt s \lt \sup S + \epsilon$

So $s$ is in the interval, $(\sup S - \epsilon, \sup S + \epsilon)$

Then $S \cap (\sup S - \epsilon, \sup S + \epsilon) \neq \emptyset$

Then $\Bigl( (\sup S - \epsilon, \sup S + \epsilon) \setminus \{\sup S\} \Bigr) \cap S \neq \emptyset$

We also know that a closed set contains all of its accumulation points and since $\sup S$ is an accumulation point of $S$, we proved $\sup S \in S$

---

Does this proof look okay? If not, can you give me hints on where I went wrong?

Thank you.

Answer 1

It is almost fine, but not entirely. You jumped from$$S\cap(\sup S-\varepsilon,\sup S+\varepsilon)\neq\emptyset$$to$$S\cap\bigl((\sup S-\varepsilon,\sup S+\varepsilon)\setminus\{\sup S\}\bigr)\neq\emptyset\tag1$$without any justification. If it turns out that $s=\sup S$, $(1)$ may well be false.

This is easy to solve, though. Just begin by saying that this is a proof by contradiction. That is, you assume that $\sup S\notin S$. Then $(1)$ would be true and you reach a contradiction when you arive that $\sup S\in S$ after all.

Answer 2

At the beginning, you better mention that your S is (non-empty) and bounded so there is a real supremum.

Otherwise, your proof is correct.

You have explained every step very clearly. You may as well make a proof based on sequences which is basically the same idea.