$A = \left\{\frac{n+m}{nm}:n,m\in \mathbb{N} \right\}$
How would I prove that A is not a compact set by using the definition of a compact set?
I do not fully understand yet how a compact set works. I probably would have to find a subsequence with its limit outside of A but I would not know how to do that.
I'll assume that $\mathbb N=\mathbb Z_{>0}$.
Finding a sequence in $A$ whose limit is outside of $A$ is a good approach. This will show that $A$ is not closed, and therefore not compact.
Try looking at the sequence $\left(\frac{n+n}{n^2}\right)_{n\geq 1}$ in $A$.
Cover this set with the open sets $(1/k,\infty)$ for natural $k$. This is an open cover and so we must find some finite subcover if this will be a compact set. However if we take any finite collection of these open sets there exists a minimum $1/k_0$ among them. Since $\frac{n+m}{nm}$ can be made arbitrarily small by fixing $n$ or $m$ then increasing the other this finite collection cannot cover the set entirely. Since no finite subcover exists for this open cover the set cannot be compact.
Don't use the definition of compact but rather a property of compact sets, the Bolzano Weierstrass theorem.
edit| this argument is wrong (thank you @PeterSzilas) Suppose for contradiction that $A$ were compact. Since the sequence $(\frac{1+n}{n})_{n \geqslant 1}$ is contained in $A$, its limit, $1$, would also be in $A$.
But that would imply that there are $m,n \in \mathbb{N}\setminus > \{0\}$ such that $1=\frac{m+n}{mn}$, ie $m+n = mn$. But then $$ m = > mn -n = n(m-1)$$ which implies that $m=0$, a contradiction (not, since $m=n=2$ is a solution).
edit| consider this sequence instead :
The sequance $\big(\frac{n+n}{n^2} \big)_{n \geqslant 0}$ is a sequence of elements of $A$, which converges to $0$, which isn't an element of $A$. By contraposition, $A$ is hence not compact.
