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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?

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    $\begingroup$ 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 $(\sup(A)-\delta,\sup(A)+\delta)\subset \mathbb{R}\setminus A$. $\endgroup$
    – IamWill
    Nov 17, 2019 at 20:49
  • $\begingroup$ Willy.K Why not write your comment as an answer? $\endgroup$ Nov 17, 2019 at 21:09

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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 $a^{\prime} \in A$.

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  • $\begingroup$ Thanks for clarifying. I wasn't sure if there was an easier quicker way $\endgroup$
    – Jac Frall
    Nov 17, 2019 at 20:51

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