$\sup$ and $\inf$ of $\{a - \frac1n|n\in \mathbb N^*\}$

Question

I have looked at similar questions on here but still cannot reach a conclusion for this particular one. I have gathered that as n increases, $\frac1n$ tends to $0$, meaning $a - \frac1n$ tends to $a$. So the set must be bounded below by $a - 1$, and bounded above by $a$.

To prove $\inf(A) = a - 1$, I started by assuming that $\inf(A) = l$ for some $l > a - 1$. So $a - l < 1$. Set $a - 1 = ε$. By Archimedean Property, there exists some $ñ > ε$. So $ñ > \frac1{\epsilon} \implies a - l > \frac1ñ$

This seems to imply $l < a - \frac1ñ$, so I did not arrive at a contradiction. Does this mean there is another possible value for the infimum? I can't think of what it could be.

Answer 1

You have to do things in order. First, manipulating inequalities to highlight a minorant and a majorant; and then prove that it's the $\inf$ and the $\sup$. $$\text{let }1\leq n$$

$$0< \frac1n \leq 1$$$$-1\leq -\frac1n<0$$

$$\color{red}a-1\leq \color{red}a-\frac1{\color{green}n}<\color{red}a$$ Then, as @hff1 and @fleablood said, $a-1=a-\frac1{\color{green}1}$ is the $\inf.$

What you still have to prove is that $$\forall \epsilon>0, a-\epsilon \text{ is not a majorant}$$to prove that $a$ is the smallest of the majorants, i.e. the $\sup.$

And the arguments you have cited are used as follows :

$$\text{Let }\epsilon >0$$$$\text{Let }\widetilde{n}>0 \text{ s.t. } \widetilde{n}\epsilon>1$$$$\epsilon>\frac1{\widetilde{n}}$$ $$-\epsilon<-\frac1{\widetilde{n}}$$$$\color{red}a-\epsilon<\color{red}a-\frac1{\widetilde{n}}$$