# Find supremum(S), infimum(S), max(s), min(S)

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

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 $$Sup(S)=\nexists$$

• 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 $\sup[0,1)=1$. Commented Oct 13, 2019 at 12:25
• True, thank you very much Commented Oct 13, 2019 at 12:30
• 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 $\sup \, S$? Commented Oct 13, 2019 at 12:31
• That is the smallest upper bound, but it does not need to belong to the set right? Commented Oct 13, 2019 at 12:42
• 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 $\frac{1}{2}(u + 1)$? Commented Oct 13, 2019 at 12:49

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., $$\sup S = 1$$.