Find the infimum of the set $S=\left\{\frac{1}{m}-\frac{1}{n} \, : m,n \in \mathbb{N^+}\right\}$

Question

I need to find the infimum of the set $$S=\left\{\frac{1}{m}-\frac{1}{n} \, : m,n \in \mathbb{N^+}\right\}$$ and formally prove that it is indeed the infimum of $S$.

From intuition, I know the $\inf S=-1$ if $m\to\infty$ and $n=1$.

However, I'm generally having trouble with formally proving supremum and infimum related questions, despite knowing their definitions. Being of a engineering background, I find it challenging to get past my intuition. I tend to go around in circles with symbols whilst trying to prove questions such as this.

So any tips on proving this problem would be appreciated.

Answer 1

Your guess is correct. To prove that, note by definition, $c= \inf S$ is the greatest lower bound. So, to show that $c = \inf S = -1$, you need to show

• $-1$ is a lower bound, and
• Any $d >-1$ is not a lower bound of $S$.

The first part is probably easy. For the second part, Just use your intuition: If $d$ is a lower bound, then put $n=1$, we get

$$\label{1}d \le \frac{1}{m} - 1\tag{1}$$

If $d>-1$, then there is $m$ large so that $d > -1 + \frac 1m$ (Not sure if you want to use the Archimedean property to justify this). This contradicts to $\ref{1}$. Thus if $d>-1$, $d$ is not a lower bound of $S$.

Answer 2

First prove: $$\forall x \in S: -1 \leq x$$ Then: $$\forall \epsilon>0, \exists x \in S: -1+ \epsilon > x$$

Answer 3

Inf$(A-B)=$ Inf$(A)$-Sup$(B)$ so $$Inf\{\frac{1}{m}-\frac{1}{n} :m,n\in\mathbb{N}\}=0-1=-1$$