Infimum and supremum of the set $A\subset \mathbb Q$

Question

Find the infimum and supremum of the set: $$A=\left\{\frac{m-n-1}{mn+4m+3n+12}:m,n\in \mathbb N\right\}$$

My mind was drifting and I entered a circle by factorizing the algebraic fraction: $\frac{m-n-1}{mn+4m+3n+12}=\frac{m-n-1}{m(n+4)+3(n+4)}=\frac{m+3-n-4}{(m+3)(n+4)}=\frac{1}{n+4}-\frac{1}{m+3}$

I thought I could fix either $m$ or $n$ and examine different cases, but I haven't come up to anything formal according to the axioms of the field $\mathbb Q$, so I started believing this is alternating without a satisfying proof. The task is in the unit before the limits. What would be a wise first step?

Answer 1

$$\lim_{n\to\infty}\frac{1}{n+4}-\frac{1}{1+3}\leq\frac{m-n-1}{mn+4m+3n+12}=\frac{1}{n+4}-\frac{1}{m+3}\leq\frac{1}{1+4}-\lim_{m\to\infty}\frac{1}{m+3}$$$$\implies \inf(A)=0-\frac{1}{4}=-\frac{1}{4},\ \sup(A)=\frac{1}{5}-0=\frac{1}{5}.$$

Remember, that the infimum of a decreasing sequence is its limit and the supremum of a increasing sequence is its limit.

Answer 2

So, now we can write $$A = \{\frac{1}{n+4} - \frac{1}{m+3}: m, n \in \mathbb{N} \}$$

Observe that the condition $m, n \in \mathbb{N} $ can be translated as $m, n\geq 1$ belonging to the set of integers.

---

We know that the infimum of $A$ (a subset of $\mathbb{Q}$) is the greatest element of $\mathbb{Q} $ less than or equal to all elements of $A$. This means we have to find the minimum value of an element in $A$ (why?). This can be found by putting $m = 1$ and $n \rightarrow \infty$, to get infimum = $-\frac14$.

---

Similarly, the supremum of $A$ of $\mathbb{Q} $ is the samleest element of $\mathbb{Q} $ larger than all elements of $A$. This means we have to find the maximum value of an element in $A$ (why?). This can be found by putting $n =1, m\rightarrow \infty$ to get supremum = $\frac15$.

Hope this helps you.