# find the supremum and infimum of the following set.

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?

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, $$\sup A = 8$$.

• Ok. Thanks. Now, why not to use direct $\sup$ definition? Aug 21, 2021 at 14:41
• @zkutch Does it similar with the infimum proof part? How to begin, since it was the union of two sets. Aug 21, 2021 at 14:45
• I wrote part of proof for $\sup$. Similar works for $\inf$, also. Aug 21, 2021 at 14:48
• 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.
– user637978
Aug 21, 2021 at 14:48

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?
• 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? Aug 21, 2021 at 15:05
• How to fix it ? It doesn't work generally, especially for the big $\epsilon \gt 0$, right? Aug 21, 2021 at 15:06
• 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 $1<1+\varepsilon$. Aug 23, 2021 at 10:47