We have a theorem $A\subseteq \mathbb{R}$ is n.w.d. $\longleftrightarrow \text{cl}(A)$ is n.w.d.
Where n.w.d. means "no-where dense" of course...
I was thinking to myself-- does there exist two n.w.d. distinct not closed sets whose closures are the same? I think this is possible with some weird discrete sets in the reals... For example:
$$
A=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}\cup \left\{2+\frac{1}{n}: n \in \mathbb{N}\right\}, \\B=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}\cup\{0\}\cup \left\{2+\frac{1}{n+1}: n \in \mathbb{N}\right\}
$$
This is an example that answers my question in the affirmative, ... But this example is really unsatisfying...
So, my question is: what are the known limitations to this? Can distinct not closed no-where dense sets whose closures are the same be both uncountable? I suspect the answer is yes, with some careful attention to details. If two uncountable sets $A,B$ are disjoint but no where dense, not closed, how big can $\text{cl}(A)\cap \text{cl}(B)$ be? Size continuum?
As a variation of the above question, what if the sets $A$, $B$ are precisely in the above question except their intersection is finite? countable?i.e. (almost disjoint)
I realize that I asked a lot of questions in one post but that is because I suspect that there exists a relatively short answer that covers a lot of my questions.
I'm not sure how to tag this question, feel free to re-tag as needed.
Easily arranged.
Just take any two subsets of the Cantor set (or any non-discrete nowhere dense set) such that the closure of each is the Cantor set. For example, if we consider the set of interval endpoints from the construction, this is a countable dense subset of the Cantor set, and its complement is also dense in the Cantor set.
Both are subsets of the Cantor set, so they are nowhere dense in $\Bbb R$. They are even disjoint.