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.
Best Answer
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.