Let $c\in\mathbb{N}$ and $S=\{(x,y):f(x,y)< c< g(x,y)\}$ where $f,g$ are some function maps from $\mathbb{R^2}$ to $\mathbb{R}$.
Can we find the Interior, boundary, and closure of $S$ ?
If not, for sets like this do they share some general property that can be helpful when we want to find their Interior, boundary, and closure?
My attempts
Here $S$ can be expressed as the intersection of two half sapces in $\mathbb{R}^2$ such that \begin{align} S^\circ=&(\{(x,y):f(x,y)< c\}\cap\{(x,y):c< g(x,y)\})^\circ\\ =&\{(x,y):f(x,y)< c\}^\circ\cap\{(x,y):c< g(x,y)\}^\circ\\ =&\{(x,y):f(x,y)< c\}\cap\{(x,y):c< g(x,y)\}=S\\ \overline{S}=&\overline{\{(x,y):f(x,y)< c\}\cap\{(x,y):c< g(x,y)\}}\\ =&\overline{\{(x,y):f(x,y)< c\}}\cap\overline{\{(x,y):c< g(x,y)\}}\\ =&\{(x,y):f(x,y)\le c\}\cap\{(x,y):c\le g(x,y)\}\\ =&\{(x,y):f(x,y)\le c\le g(x,y)\}\\ \partial S=&\overline{S}\setminus S^\circ\\ =&\{(x,y):f(x,y)\le c\le g(x,y)\}\cap\{(x,y):f(x,y)< c< g(x,y)\}^c\\ =&\{(x,y):f(x,y)\le c\le g(x,y)\text{ and }(f(x,y)\ge c\text{ or } c\ge g(x,y))\}\\ =&\{(x,y):f(x,y)\le c\le g(x,y)\\ &\text{ and }(f(x,y)< c\text{ implies } c\ge g(x,y))\\ &\text{ and }(c<g(x,y)\text{ implies }f(x,y)\ge c)\} \end{align}
Consider four cases
$1.f(x,y)< c< g(x,y)$ then $c\ge g(x,y)$ contradiction.
$2.f(x,y)< c= g(x,y)$ which is fine.
$3.f(x,y)= c< g(x,y)$ which is fine.
$4.f(x,y)= c= g(x,y)$ which is fine
Hence we have $$\partial S=\{(x,y):f(x,y)< c= g(x,y)\}\cup\{(x,y):f(x,y)= c< g(x,y)\}\cup\{(x,y):f(x,y)= c= g(x,y)\}$$ This solution intuitively seems make sense to me $\dots$
If this is correct, would this solution also has a generalization to $\mathbb{R}^n$, such that
$$S=\{(x_1,\dots,x_n):f(x_1,\dots,x_n)<c<g(x_1,\dots,x_n)\}$$
And my another confusion about this is, say we have $$S_1=\{(x,y):f(x,y)\le c\le g(x,y)\}$$ $$S_2=\{(x,y):f(x,y)\le c< g(x,y)\}$$ $$S_3=\{(x,y):f(x,y)< c\le g(x,y)\}$$ they also share the same Interior,closure and boundary as $S$, is this correct?
Thanks for your help.