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

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

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

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

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.

• Thank you. My answer why we should find the minimum & maximum of A is to check whether infimum & supremum exist in $\mathbb Q$ and if so, then the greatest minorant and the least majorant are in $\mathbb Q$ – Praskovya2.718281828 Nov 3 at 12:38