Supremum and infimum of $\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$

Solution 1:

• $A=B-B$ where $B=\{\frac{1}{n} : n \in \mathbb{N}\}$ and $X-Y=\{x-y: x \in X, y \in Y\}$.

• $\inf B=0$, $\sup B=1$. ($\inf B=0$ is the Archimedean property.)

• $\inf (X-Y) = \inf X - \sup Y$ implies $\inf A = \inf B - \sup B = 0 - 1 = -1$.
• $\sup (X-Y) = \sup X - \inf Y$ implies $\sup A = \sup B - \inf B = 1 - 0 = 1$.

Solution 2:

I don't believe you are using the archimedean property in a meaningful way. The idea that there exists a number $x$ with the properties that you describe is trivial, take $x=1$ and $n'=2$. That doesn't help with your proof.

What is more meaningful is that for all positive real numbers $x$ we can find an $n'\in\mathbb N$ (based on $x$) such that $1<n' x$. Since your proof doesn't make this statement, most of what comes after your use of the archimedean property needs some tweaking.

You do successfully prove that $-1$ is a lower bound. If I were approaching the proof (and it greatly depends on what theorems you have already proven) I would suggest assuming that $y$ is a greater lower bound than $-1$ and deduce that there must be an element of $A$ between $-1$ and $y$, which is a contradiction.

Edit, in response to OP's edit: Your proof actually makes no use of the archimedean property anyway. You invoke it to create this number $-1+x$, but the proof makes no real use of this number anyway. You can do without all of that, just assuming that $y$ is a greater lower bound than $-1$, so $-1<y$ and $\forall x \in A,\, y\le x$.

Now you try to prove that there must be an element of $A$ less than $y$ (which I will leave to you to do). Note that it is not enough to just show that there is some rational number between $-1$ and $y$, the elements of $A$ take a particular form.