The supremum and infimum

Question

Let $S=\{5/n \mid n\in\mathbb N\}$. Find $\sup(S)$ and $\inf(S)$.

I can clearly see that $\sup(S)=5$ and $\inf(S)=0$.

I know that the definition of supremum is for all $\gamma <5$, there exists $x\in S$ such that $\gamma\leq x$. I also know that the definition of infimum is for all $\gamma>0$, there exists $x\in S$ such that $\gamma\geq x$.

I am not allowed to use the Archimedian Property (haven't been taught it in class), but I can use the Completeness Axiom which states that every nonempty set that has an upper bound must have a least upper bound. Clearly, S is not empty and 5 is an upper bound for S, so by Completeness Axiom, S must have a supremum.

I'm unsure of how to continue the proof.

Answer 1

To show that $5$ is the upper bound, show that each successive term is decreasing (i.e. $\frac{5}{n+1}< \frac5n$). So no term exceeds $5$, so $5$ is the least upper bound.

To show that $0$ is the greatest lower bound, first you know that $0$ is a lower bound since you're dividing positive numbers so $5/n$ can never be negative. Then show that for any $\epsilon>0$, you can find some $n$ such that $5/n<\epsilon$. So $0$ is the greatest lower bound.

Answer 2

It is not hard to show that $\sup S = 5$. To show that the sequence $\{5/n\}_{n \ge 1}$ is decreasing, one uses the fact that if $a > b > 0$, then $1 > b/a > 0$, hence $1/b > 1/a > 0$, thus $5/b > 5/a > 0$.

Proving the infimum is $0$ is essentially equivalent to a proof of the Archimedean property itself. Consequently, you should look for such a proof and adapt it to this particular case.