All I have found are various definitions of the clousure of a set. The only solving strategy I found was to intersect all closed subsets containing the subset whose closure I am looking for. However this does not help me out in my exercise in which I am asked to find the closure of $A$:
$$ A:=(Q \cap (0,\infty)) \setminus\left\{n^{-1} \mid n \in \mathbb{N} \right\} \text{ in } (0,\infty) $$
I have literally no idea how to approach this problem and did not find anything specific on the internet. Can someone please help me solving this?
Let $x \in (0, +\infty)$. Consider the sequence $$u_n = \frac{\lfloor 10^nx \rfloor}{10^n}$$
where $\lfloor. \rfloor$ denotes the floor function. You have $$x- \frac{1}{10^n} =\frac{10^nx-1}{10^n} \leq u_n \leq \frac{10^nx}{x} = x$$
so by comparison the sequence $(u_n)$ converges to $x$.
Now consider, for each $k$, consider the set $$A_k = \left\lbrace n \in \mathbb{N}^* \quad | \quad \frac{1}{n} \in \left(u_k, u_k + \frac{1}{k} \right) \right\rbrace $$
This set is a subset of $\mathbb{N}^*$, and therefore is either empty, either has a minimal element.
If $A_k$ is empty, let's define $v_k = u_k$. If $A_k$ has a minimal element $n_0$, then define $$v_k = \max \left( \frac{u_k + \frac{1}{n_0}}{2}, \frac{\frac{1}{n_0+1} + \frac{1}{n_0}}{2} \right)$$
I let you check that the sequence $(v_k)$ tends to $x$ and is composed of elements of $A$.
This would prove that the closure of $A$ is the entire $(0, +\infty)$.