Prove that infimum (A)=0 and that supremum (A)=1 in the following set

Question

$$A=\{\frac{n}{m}:m,n \in \mathbb{Z}^+, m>n\}$$

Now, I know that, as $n$ approaches $0$ from above and as $m$ approaches infinity, $\frac{n}{m}$ gets arbitrarily close to $0$, but my professor doesn't accept waffly terms like 'arbitrarily close' as we haven't rigourously defined such terms yet. So, how would I go about proving these two facts formally?

Answer 1

For every $x=\frac{n}{m} \in A$ we have $m>n$. Therefore $0<x<1$ for every $x \in A$. It follows that $\inf A\ge 0$ and $\sup A \le 1$.

Let $\varepsilon>0$ be sufficiently small. For every $\frac{n}{m} \in A$ we have $$ \frac{n}{m} \ge \frac{1}{m} \in A. $$ Since $\lim_{m \to \infty}\frac{1}{m}=0$, there is an $m_0=m_0(\varepsilon)\in \mathbb{N}$ such that $$ A\ni \frac{1}{m}< \varepsilon \quad \forall m >m_0 $$ Thus $\inf A=0$.

Notice that $m>n$ for every $\frac{n}{m} \in A$. Therefore $m \ge n+1$ (because $m,n \in \mathbb{N}$), and so we have $$ \frac{n}{m}\le \frac{n}{n+1} \quad \forall \frac{n}{m}\in A. $$ But $\lim_{n\to \infty}\frac{n}{n+1}=1$, i.e. there is some $n_0=n_0(\varepsilon) \in \mathbb{N}$ such that $$ -\varepsilon\le \frac{n}{n+1}-1 \le \varepsilon \quad \forall n >n_0, $$ in particular we have $$ A\ni \frac{n}{n+1}\ge 1-\varepsilon \quad \forall n >n_0. $$ Hence $\sup A=1$.

Answer 2

I think you can see that

$0< \frac{n}{m} < 1$

Now you can't say "as n approaches 0" because $n \in \mathbb{Z}_+$.. but you can say that if $n=1$, then $\frac{n}{m} = \frac{1}{m}< \epsilon$ for every $\epsilon > 0$ (just take $m > \frac{1}{\epsilon}$)

So you will never reach $0$, but for every number $\epsilon$ as close to zero as one wants, then we can find $n, m$ such that $\frac{n}{m} < \epsilon$. This implies $\inf(A) = 0$.

In fact, if you defined $\inf(A)$ in the standard way, (as the greatest element of A that is less than or equal to all elements), then one can say: "Suppose $\inf(A) = \epsilon \neq 0$. This means $\epsilon \le x, \forall x \in A$, but we have shown this is not the case.

So $\epsilon \le 0$.
We have already shown that if $\epsilon \le 0$, then it's true $\epsilon \le x, \forall x \in A$.

What is then the greates $\epsilon$ with such a property? $0$ ! So $\inf(A) = 0$

Can you complete the proof and show that $\sup(A)= 1$?

EDIT

I see maybe you need a clearer explanation. We want to prove $\inf(A) = 0$

This means that we want to prove that $0 \le x \ \forall x \in A$, and that $0$ is the greatest number with such property.

We're gonna do it with the so called Proof by contradiction (http://en.wikipedia.org/wiki/Proof_by_contradiction)

Let's suppose $\inf(A) = b > 0$. ($b < 1$) Then $b \le \frac{n}{m} \forall \ n, m$ (1)

For $n = 1$, $m = \frac{2}{b}$ (note that this way $\frac{n}{m} \in A$) we have $b \le \frac{b}{2}$ which is false. So $b$ cannot be $> 0$ and has to be $b = 0$.

Basically you start with the definition of $\inf(A)$ and try to find a contradiction.

Answer 3

Start from the definition of $\inf (A)$. When you say $x=\inf (A)$ you asserting that $x \le$ every member of $A$ and that no number greater than $x$ is also $\le$ every member of $A$. It sounds like you have the first half. For the second, if $y \gt 0$ you have to find a member of $A$ that is $\lt y$. The $\sup$ argument is symmetric.