I have little trouble with the following examples of sets to find its minimum, maximum, supremum and infimum. Sometimes it gets tricky if the example is not trivial. I know that here are many examples like that but I want an answer for those two. For this examples I will show my way of thinking, correct me please:

  1. $ x = \big\{x:x=\frac{n-1}{n},n \in N^+\big\} $

  2. $S=[0,1)\in R $

I'm using the following definitions that supremum is the smallest upper bound $x \in R$, and s is the element of a set S , $\forall \epsilon >0 , \exists s \in S:s>x-\epsilon , $ respectively the infimum is the greatest lower bound $\forall \epsilon >0 , \exists s \in S:s<x+\epsilon , $

  1. for $ n=1,2,3,4... , x=0,\frac{1}{2}, \frac{2}{3}, \frac{3}{4}... $ the values of x will never reach 1

So from my point of view, $min(S)=0$ and $ Inf (S)=0$ but $max(S)=\nexists $ and I think that $sup(S)$ does not exist, why I think that? From the definition, let us pick a number $s $ that would be greater than least upper bound minus a very small number but we don't know what is the smallest upper bound because the set is infinite

2.$min(S)=0$ and $max(S)=\nexists$ $Inf (S)=0$ but what with $Sup(S)=?$

If we pick a number $s=1 \notin S $ that for sure $1>x-\epsilon $ but the element s is not in the set so how to prove that for example $0.9999999>1-0.0000001$ , easily they gonna be equal but is it possible the left-hand side of the inequality to be grater or simply $Sup(S)=\nexists$

  • 3
    $\begingroup$ Is $1$ an upperbound of $[0,1)$? Can you find an element smaller than $1$ that is an upperbound of $[0,1)$? If you have answered the first question with "yes" and the second with "no" then you have confirmed that $1$ is the smallest upperbound of $[0,1)$ right? And can state in good conscience that $\sup[0,1)=1$. $\endgroup$
    – drhab
    Oct 13, 2019 at 12:25
  • $\begingroup$ True, thank you very much $\endgroup$
    – Michael W
    Oct 13, 2019 at 12:30
  • $\begingroup$ In both both cases the supremum is $1$. I'd advise you work through the definion of infimum and supremum once again. The supremum is the smallest upper bound of the set. For $S=[0;1)$ clearly $[1;\infty)$ are all the upper bounds (every number smaller than $1$ is either in $S$ or even below $S$ - so a lower bound). What does that say about $\sup \, S$? $\endgroup$
    – C. Brendel
    Oct 13, 2019 at 12:31
  • $\begingroup$ That is the smallest upper bound, but it does not need to belong to the set right? $\endgroup$
    – Michael W
    Oct 13, 2019 at 12:42
  • $\begingroup$ A bounded set always has a supremum, this is the axiom you can add to the axioms of $\mathbb{Q}$ to make it into $\mathbb{R}$. I'd stick to the definition that the supremum is the smallest upper bound. If $u < 1$ is a potential upper bound, what about $\frac{1}{2}(u + 1)$? $\endgroup$
    – user388557
    Oct 13, 2019 at 12:49

1 Answer 1


The statement "Every non-empty set $S \subseteq \mathbb{R}$ which is bounded from above has a least upper bound" is often stated as an axiom when defining the set $\mathbb{R}$ of real numbers. This supremum property is equivalent to completeness, i.e., every Cauchy sequence is convergent. Regardless, once this axiom is accepted, then every non-empty set $S \subseteq \mathbb{R}$ which is bounded from above has a least upper bound.

Turning to the task at hand. With $S = \{ 1 - \frac{1}{n} \: : \: n \in \mathbb{N} \}$ it is clear that $r= 1$ is an upper bound. We claim that $r=1$ is the least upper bound. To that end, let $\epsilon > 0$ be given. We must show that $t = 1 -\epsilon$ is not an upper bound. This means finding $n \in \mathbb{N}$ such that such that $1 - \epsilon < 1 - \frac{1}{n}$ or equivalently $\epsilon^{-1} < n$. This is easily done. We can choose $n = \lceil \epsilon^{-1} \rceil + 1$. In short, $r=1$ is an upper bound for $S$ and any number strictly smaller than $r=1$ is not an upper bound for $S$. It follows that $r=1$ is the least upper bound for $S$, i.e., $\sup S = 1$.


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