Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$\inf(A)=-\sup(-A).$$

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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 $\sup A = \sup B - \inf B = 1 - 0 = 1$.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.