Prove that closed bounded set contains supremum

Prove that a closed bounded set contains its supremum

Proof:

Consider a closed bounded set $$A$$. Suppose to the contrary that $$\sup (A) \notin A$$. We know that since $$A$$ is closed that $$\mathbb{R} \setminus A$$ is open. Then since $$\sup (A)$$ is an element of $$\mathbb{R}\setminus A$$ it follows that we can find a $$\delta >0$$ such that $$(\sup (A) - \delta,\; \sup (A) + \delta) \subset \mathbb{R}\setminus A$$. Then $$\sup(A) -\delta$$ is clearly an upper bound of $$A$$ which is a contradiction.

The one thing that I think I may be missing is that I need to show that $$\sup(A)-\delta$$ is an upper bound. How would I do that in this case?

• You could change a little bit your argument: by the definition of supremum, for every $\delta >0$ fixed there is an $x \in A$ such that $\sup(A)-\delta \le x$, so it cannot be $(\sup(A)-\delta,\sup(A)+\delta)\subset \mathbb{R}\setminus A$. Commented Nov 17, 2019 at 20:49
• Willy.K Why not write your comment as an answer? Commented Nov 17, 2019 at 21:09

1 Answer

If possible suppose $${\mathrm{sup}}(A) - \delta$$ is not an upper bound of $$A$$. Then there must be $$a^{\prime} \in A$$ so that $$a^{\prime} > {\mathrm{sup}}(A) - \delta$$. Since $$a^{\prime} \in A$$, we have $${\mathrm{sup}}(A) - \delta < a^{\prime} \leq {\mathrm{sup}}(A) < {\mathrm{sup}}(A) + \delta$$. This implies $$a^{\prime} \in ({\mathrm{sup}}(A) - \delta, {\mathrm{sup}}(A) + \delta) \subseteq {\mathbb{R}} \setminus A$$. This contradicts $$a^{\prime} \in A$$.

• Thanks for clarifying. I wasn't sure if there was an easier quicker way Commented Nov 17, 2019 at 20:51