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

I would like to verify my proof of the following:

Let $A=\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$. I want to show that $-1$ and $1$ are the infimum and supremum respectively.

First I will show that $-1$ is the infimum. To show this I will first demonstrate that it is indeed a lower bound. Observe: $$\frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-1 \geq -1.$$

I now claim that $-1$ is the greatest lower bound. So, by the archimedean property, there exists an $x \in \mathbb{R}$, $x>0$ and $n' \in \mathbb{N}$ such that $1<n'x$. Thus, $\frac{1}{n'}<x$ and $$\frac{1}{n'}-1<x-1.$$ From the above it is clear that $\frac{1}{n'}-1 \in A$. Also, it is clear that $-1<-1+x$.

Let $y=\inf(A)$. So we let $$-1<y\leq-1+x.$$ By completeness there exists an $r\in\mathbb{Q}$ such that $$-1<r<y\leq-1+x.$$ Thus $y$ is not the infimum.

I was going to note that the supremum was 1 through the relationship between $$\inf(A)=-\sup(-A).$$

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Do you mean A has only one element (which is what you've written), or do you actually mean $A = \{1/n - 1/m \mid n,m\in\mathbb{N}\}$? – JeffE Sep 14 '12 at 1:12
Yes, I did mistype the set. Thanks. – emka Sep 14 '12 at 1:13
– Martin Sleziak Jun 21 '15 at 9:05

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.

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I corrected the direction of those symbols. That was just an error on my part translating my handwritten notes into \LaTeX. After correcting those symbols, is this proof correct. I did find an r that is between -1 and y. – emka Sep 14 '12 at 1:58
Ah, I see that. The statement invoking the archimedean property needs some work still. – process91 Sep 14 '12 at 1:59
Could you elaborate on that? I guess I'm not sure how I'm misusing the archimedean property. – emka Sep 14 '12 at 2:03
@EMKA The archimedean property makes a statement about all positive real numbers. Your statement is that there exists some particular real number. I have updated my answer to reflect your edit. – process91 Sep 14 '12 at 2:08
• $A=B-B$ 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 $\sup A = \sup B - \inf B = 1 - 0 = 1$.
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