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Let $A = [1,3) \cup (5,8)$. Find the supremum and infimum of $A$.

Firstly, I claim that $\sup A = 8$ and $\inf A = 1$. Now, by definition, $8$ is the upper bound of $A$. My next step is use this theorem, but I think there is a simple proof of this question without any theorem:

Let $A$ and $B$ be bounded subsets of $\Bbb R$. Then, $\sup (A \cup B) = \sup\{\sup A, \sup B \}$ is holds.

That is, I divide the set $A$ into two sets, say $A_1 = [1,3)$ and $A_2 = (5,8)$, then I prove that $\sup A_1 = 3$ and $\sup A_2 = 8$, and then use the above theorem, and done.

But, my question is, does there exist a simple proof that without any theorem? How about the infimum proof? Does there exist theorem that says a similar argument same as the supremum theorem above?

Thanks in advanced.

EDIT: I've did the approach of proof of supremum as @zkutch post below. Please correct me if I wrong. Here it is:

Claim that $\sup A = 8$. By definition, $8$ is the upper bound of $A$. We want to prove that $8$ is the least upper bound of $A$. Suppose for contradiction, that $8$ is not the least upper bound of $A$. Then, there exist a number $\epsilon \gt 0$ such that $8-\epsilon$ is also an upper bound of $A$. To contradict this, we exhibit $x \in A$ such that $8-\epsilon \lt x \lt 8$. Since $0 \lt \frac{\epsilon}{2} \lt \epsilon$, we can see that $x = 8 - \frac{\epsilon}{2}$ satisfies $8-\epsilon \lt x \lt 8$.

Now, since $8-\epsilon$ is (by assumption) a upper bound of $A$ and $5+\epsilon \in A$, then we have $5+\epsilon \le 8-\epsilon$, showing that $x \in A$. Thus, $8-\epsilon$ is not an upper bound of $A$, a contradiction. Hence, $\sup A = 8$.

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  • $\begingroup$ Ok. Thanks. Now, why not to use direct $\sup$ definition? $\endgroup$
    – zkutch
    Commented Aug 21, 2021 at 14:41
  • $\begingroup$ @zkutch Does it similar with the infimum proof part? How to begin, since it was the union of two sets. $\endgroup$
    – lap lapan
    Commented Aug 21, 2021 at 14:45
  • $\begingroup$ I wrote part of proof for $\sup$. Similar works for $\inf$, also. $\endgroup$
    – zkutch
    Commented Aug 21, 2021 at 14:48
  • $\begingroup$ by definition, 8 is AN upper bound of $A$. Upper bounds are not in general unique. The supreme of $A$ on the other hand, defined as the least upper bound, is unique. $\endgroup$
    – user637978
    Commented Aug 21, 2021 at 14:48

1 Answer 1

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Definition of $\sup$ requires two property: first there is, because $8$ is more, then any number in union. Then, for every $\varepsilon >0$ is possible to find number from $(5,8)$, which is more then $8-\varepsilon $. Can you do it?

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  • $\begingroup$ Yes, I've did it. Please check my approach below and please wait, thanks. :) $\endgroup$
    – lap lapan
    Commented Aug 21, 2021 at 14:50
  • $\begingroup$ Please reload the page and see my edit above. $\endgroup$
    – lap lapan
    Commented Aug 21, 2021 at 15:02
  • $\begingroup$ Yes, I see. You need warranty, that $x=8-\frac{\varepsilon}{2}>5$, but, generally, there is not. Do you see how to fix this moment? $\endgroup$
    – zkutch
    Commented Aug 21, 2021 at 15:05
  • $\begingroup$ How to fix it ? It doesn't work generally, especially for the big $\epsilon \gt 0$, right? $\endgroup$
    – lap lapan
    Commented Aug 21, 2021 at 15:06
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    $\begingroup$ We use that $(5,8) \subset A$. So, if we find $x$ for second property of $\sup$ in $(5,8)$, then, we find it in $A$. We exploit that $A$ have simple structure. And, yes, for $\inf$ works same way, but here it's more easy: we have $1\in A$, so second property is obvious $1<1+\varepsilon$. $\endgroup$
    – zkutch
    Commented Aug 23, 2021 at 10:47

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