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Does a nonempty closed and bounded set in $\mathbb{R}$ necessarily contain its supremum and infimum?

My thought process: Closed and bounded means compact in $\mathbb{R}$, and a continuous function on a compact set admits a minimum and maximum. The identity function is continuous, so it admits a minimum and maximum on any compact set, which must be the minimum and maximum of the compact set.

Is this valid? Is there a simpler proof/disproof?

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    $\begingroup$ Your proof is valid. You could also just construct sequences in the set converging to the inf and sup. Since the set is closed, the limits will be in the set. Then the inf and sup are in the set. $\endgroup$
    – user12014
    Oct 25 '11 at 2:50
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The proof is correct, but it's rather non-minimal, since the extreme value theorem you use is stronger than the result you want and the proof for this specific case is easier than the proof of that theorem.

Assume the set didn't contain its supremum (which exists since the set is bounded). Since the set is closed, an entire neighbourhood of the supremum lies outside the set. This neighbourhood contains upper bounds of the set that are lower than the supremum, which is a contradiction.

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  • $\begingroup$ You might want to explain why you can separate the closed set from its supremum, this is because $\mathbb{R}$ is a regular space. $\endgroup$ May 28 '18 at 14:23
  • $\begingroup$ @JensWagemaker: I don't understand. I wasn't separating the closed set from its supremum. I only used the fact that a neighbourhood of the supremum is disjoint from the set; I didn't use the stronger fact that a neighbourhood of the supremum is disjoint from an open set containing the set, which is what regularity would yield. (Assuming we're talking about the same kind of regularity (youtu.be/3kZUWkB7eZE?t=47s) :-). Or am I missing something? $\endgroup$
    – joriki
    Jun 8 '18 at 5:58
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    $\begingroup$ I see, you are using something weaker than regular indeed. Does the property you use have a name? (i.e. for every closed set $S$, and $x \not \in S$, we can find an open neighborhood of $x$ that does not intersect $S$). $\endgroup$ Jun 8 '18 at 9:35
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    $\begingroup$ @JensWagemaker: Yes, that property does have a name -- it's called "closed" :-). A closed set is by definition the complement of an open set, an open is by definition an open neighbourhood of all its points, and the complement of a set by definition doesn't intersect the set. We don't need any further topological separation properties of the space for that. $\endgroup$
    – joriki
    Jun 8 '18 at 14:09
  • $\begingroup$ perfect, that settled it! $\endgroup$ Jun 8 '18 at 16:08
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Here is another way of proving the result, which is (was) similar to joriki's answer.

Let $A \subset \mathbb{R}$ be closed and bounded. If $A$ is a singleton, then we are done. As long as $A$ has a finite cardinality this argument holds. If $A$ has infinitely many elements, let $L=\sup A$, which exists since $A$ is compact and $\mathbb{R}$ is complete. The idea now is to construct a sequence in $A$ converging to $L$, from whence the claim would then follow. Take an $\epsilon > 0$, then there exists an $x_\epsilon \in A$ such that

$$\tag{1} L-\epsilon \le \ x_\epsilon \le L $$

by definition of the supremum. Since $\epsilon$ is arbitrary, we can take the limit as it approaches 0.

We would need to consider

$$ \underset{\epsilon \to 0}{\lim}\ L-\epsilon \le \underset{\epsilon \to 0}{\lim}\ x_\epsilon \le L $$ We argue that there exists an $x_\epsilon$ which still belongs to $A$ for each $\epsilon$ chosen by asserting equation (1).

This gives us $$L\le x_0 \leq L \quad \text{where} \quad x_0= \underset{\epsilon \to 0}{\lim}\ x_\epsilon $$ Thus $$L \in A$$ Namely, $A$ contains $L=x_0$ since it is a limit point, and $A$ is closed.

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  • $\begingroup$ a) $\mathbb R$ is not bounded. b) There need not exist an $x\lt\sup A$; A might consists only of a single element. c) If you take the limit of the first displayed inequality as $\epsilon$ approaches $0$, the $\lt$ would turn to $\le$, but worse, the choice of $x$ depended on $\epsilon$, so you can't just let it stand as a free variable, you need $\lim_{\epsilon\to0}x$ there; you'd have to argue that that exists, and even then it's not immediately clear why $L\in A$ should follow, since that limit need not be in $A$. $\endgroup$
    – joriki
    Oct 25 '11 at 7:27
  • $\begingroup$ @joriki. You're right on the last two counts. Perhaps there is a misunderstanding with the first one. I meant that $A$ is closed and bounded when I said $A \subset \mathbb{R}$. Otherwise my reasoning has been shown up. I've tried to improve the proof--do you think this approach will work? $\endgroup$
    – Samuel Tan
    Oct 25 '11 at 9:43
  • $\begingroup$ I don't see any changes in your answer -- perhaps they haven't percolated through the system yet? Regarding the first point, I didn't mean the first occurrence of "bounded", which I did understand to refer to $A$, but the second one, which unambiguously refers to $\mathbb R$. $\endgroup$
    – joriki
    Oct 25 '11 at 9:47
  • $\begingroup$ Oh...mea culpa. You are right on all three counts. My mistake has been rectified. Does the above now hold water? $\endgroup$
    – Samuel Tan
    Oct 25 '11 at 10:03
  • $\begingroup$ I think it's mostly OK now, except a) the $\lt$ should be $\le$ already one further up; b) it's a bit imprecise to speak of a "sequence" and then let $\epsilon$ go to $0$ and not $n$ to $\infty$, though of course the two are equivalent, and c) it is now even less the case than before that this is "similar to joriki['s answer]" :-) $\endgroup$
    – joriki
    Oct 25 '11 at 10:14

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