I was wondering if what I have done is correct.
Show that the set is not compact using the open cover definition (i.e., a subset $A$ of a metric space $\mathbb{X}$ is compact if every open cover of $A$ has a finite subcover).
$$A=\left\{\left(n, \frac{1}{n}\right):n\in\mathbb{N}\right\}\subseteq\mathbb{R^{2}}$$
$A$ is an open cover for itself. Suppose that the finite subcover $$B=\bigcup_{n=1}^{k}\left\{\left(n,\frac{1}{n}\right)\right\}\subseteq\mathbb{R^{2}}$$ contains $A$. But since the point $\left(k+1,\frac{1}{k+1}\right)\in A$ and the point $\left(k+1,\frac{1}{k+1}\right)\notin B$, then $A\nsubseteq B$, so B is an open cover without a finite subcover of A. Thus, A is not compact.
If it is correct, it seems like using open countable sets (of only one variable?) as their own open covers is the easiest way to do these problems. But, I also haven't thought of any examples where this wouldn't work or would be harder, so if someone could provide examples, then I would appreciate it.
Best Answer
In the case of $\mathbb{R}^2$, we have that $A$ is a collection of points. Here you would argue that $A$ is not compact by considering the open cover $B(a_n, 1)$, where $a_n =(n, \frac{1}{n}) \in \mathbb{R}^2$ $(n=1, 2, \ldots)$.