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Prove that if there is $A\subseteq \mathbb{R}$ not empty and bounded so $infA\leq SupA$ and infA=SupA iff A={One element}.

By definition every bounded set in $\mathbb{R}$ has Inf and Sup, therefore for every $a \in A$ $InfA \leq a \leq SupA\rightarrow InfA\leq SupA$
If A={One element} so A is both Inf and Sup but I can not see why

Git Gud
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gbox
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2 Answers2

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As you have seen, $a\in A$ implies $\inf A\le a\le \sup A$. Next, if $A$ is not a singleton, say $a,b\in A$ with (wlog.) $a<b$, then $\inf A\le a<b\le \sup A$, i.e., $\inf A<\sup A$. We conclude that $\inf A=\sup A$ can only hold if $A$ is a singleton set. As $A=\{a\}$ implies $\inf A=a=\sup A$, the proof of if and only if is complete.

Hagen von Eitzen
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  • I still can not see why if A is a singleton then infA=supA. You proved that $p\rightarrow q$ by proving $\neg q\rightarrow \neg P$? – gbox Dec 23 '14 at 10:52
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    If $A={a}$ then $a\le x$ for all $x\in A$, hence $\inf A\ge a$. On the other hand, $s\le x$ for all $x\in A$ implies $s\le a$, hence $\inf A\le a$. Similar for $\sup A$. – Hagen von Eitzen Dec 26 '14 at 12:07
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$$\inf A \le a \le \sup A$$ $$\wedge$$ $$\inf A = \sup A$$ $$\to \inf A = a = \sup A$$

That's for the 'only if. 'if' is obvious.

QED

Remark: We don't have $\inf A \le a \le \sup A$ for $A=\emptyset$

BCLC
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