How do I show this set is not compact? $A = \left\{\frac{n+m}{nm}:n,m\in \mathbb{N} \right\}$

Question

$A = \left\{\frac{n+m}{nm}:n,m\in \mathbb{N} \right\}$

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.

Answer 1

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 $A$.

Answer 2

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.

Answer 3

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.