# The supremum and infimum

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.

• I think this $S$ is $\{5,5/2,5/3,5/4,1,5/6,5/7,\ldots\}$. May 1, 2019 at 3:01

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.
• @MattSchiff No. You know that $n+1>n$ for any $n$. So flip both sides and multiply by $5$ and you've shown it. May 2, 2019 at 0:41
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.