# Supremum and Infimum using Archemedian Property

I'm having trouble understanding where the Archemedian Property can be applied when evaluating the supremum and infimum of a set S.

For example:

Suppose $S = \{\frac{1}{n} - \frac{1}{m} : n,m \in \mathbb N\}$ Find $sup(S)$ and $inf(S)$

I want to show that $sup(S) = 1$ and $inf(S) = - 1$

$sup(S) = 1$ :

First note that $\frac{1}{n} - \frac{1}{m} \leq 1 - \frac{1}{m} < 1$

Then 1 is an upper bound for $\frac{1}{n} - \frac{1}{m}$.

At this point I believe I need to apply the Archemedian Property but I'm not sure how.

I think I need to use the corollary that states : For any arbitrary $\epsilon > 0$, there exists $n_\epsilon \in \mathbb{N}$ such that $0 < \frac{1}{n_\epsilon} <\epsilon$

Could anyone give me a hint?

$N=\left\{\dfrac{1}{n} \mid n\in \mathbb{N}\right\}$

$(-M)=\left\{-\dfrac{1}{m} \mid m\in \mathbb{N}\right\}$

$(N-M)=\left\{\dfrac{1}{n}-\dfrac{1}{m} \mid n,m\in \mathbb{N}\right\}$

$\sup (N-M)=\sup N+\sup(-M)=\sup N-\inf M=1$

$\inf (N-M)=\inf N+\inf(-M)=\inf N-\sup M=-1$

Edit: You've shown $1$ is an upper bound of $S$. To show $1$ is the $least$ upper bound (i.e. supremum) of $S$, you have to show every number less than $1$ is not an upper bound. So for all $1-\epsilon$, with $\epsilon>0$, you need to show these is some element of $S$ bigger than $1-\epsilon$, meaning for some $n,m$ $$\frac1n-\frac1m>1-\epsilon$$ Choosing $n=1$ allows and subtracting both sides from one should reveal how the corollary you mentioned is useful.

• Sorry but where did $\frac{1}{n} - \frac{1}{m} > 1 - \epsilon$ come from? Oct 15, 2014 at 4:42
• @AlexWong See the edit I just made. Oct 15, 2014 at 5:06
• This leaves us with $\frac{1}{m} > \epsilon$ right? Doesn't the corollary state that we should having something of the form $\frac{1}{n} < \epsilon$? Oct 15, 2014 at 5:21
• When you have $1-\frac1m>1-\epsilon$, then this implies $\frac1m<\epsilon$. When you subtract from 1, inequalities switch. $\frac23>\frac12$, but $1-\frac23<1-\frac12$. Oct 15, 2014 at 5:36