Is the topological boundary of a set equal to the boundary of its closure or of its interior? Can you give me counter examples?
1-) Boundary of the set $S$ is the same as the boundary of the closure of $S$. I couldn't find a counterexample which would make this statement false, all my examples support this statement.
2-) Boundary of the set $S$ is the same as the boundary of the interior of $S$. What is meant by the boundary of the interior of $S$?. If $S$ is an open set then $S$ would be equal to its interior. Then this statement is the same as the 1st one.
Any help will be appreciated.
 A: Remember that $\partial S = \overline S \cap \overline C_S$, where $C_S$ is the complementary of $S$. Let $S = \Bbb Q$ with the usual topology induced from $\Bbb R$. Let $\Bbb I = C_ {\Bbb Q}$, the irrational numbers.
1) On the one hand we have
$$\partial S = \overline {\Bbb Q} \cap \overline {\Bbb I} = \Bbb R \cap \Bbb R = \Bbb R .$$
On the other hand, since $\overline S = \overline {\Bbb Q} = \Bbb R$, we have that
$$\partial \overline S = \partial \Bbb R = \overline {\Bbb R} \cap \overline {C_{\Bbb R}} = \overline {\Bbb R} \cap \overline \emptyset = \Bbb R \cap \emptyset = \emptyset ,$$
which shows that $\partial S \ne \partial \overline S$.
2) We have that $\mathring S = \mathring {\Bbb Q} = \emptyset$, therefore
$$\partial \mathring S = \partial \emptyset = \overline \emptyset \cap \overline {C_\emptyset} = \emptyset \cap \overline {\Bbb R} = \emptyset \cap \Bbb R = \emptyset .$$
Since we had already obtained $\partial S = \Bbb R$, we see that $\partial S \ne \partial \mathring S$.
