Given $S \subset \Bbb{R}$, show $\textbf{int}(S)+\textbf{ext}(S)+\partial S =\Bbb{R}$ The way I proved it is that we knwo R is open so intR=R.
For any point in IntS is inside of IntR and any point in ExtS is inside of IntR.
any point that is neither intS nor extS is still inside of IntR
So intR is collection of all intS,extS,and boundary of S, which means R is union of intS extS and boundary of S
It seems like my proof is not complete, and cant be sure whether i can call it a proof,,
Can anyone help me ?
 A: Since $\text{ext}(S), \space \text{int}(S), \space \partial S \subseteq \Bbb{R}$ it follows that the union $$\text{ext}(S) \cup \text{int}(S) \cup  \partial S \subseteq \Bbb{R}$$ Now you need to show the reverse inclusion. Recall: $$\text{ext}(S) = \Bbb{R} \setminus \overline{S} \\ \text{int}(S) =\Bbb{R} \setminus \overline{\text{ext}(S)} \\ \text{and} \quad \partial S = \overline{S}\setminus \text{int}(S)$$ Now let $x \in \Bbb{R}$. It should be clear that $x \in \overline{S}$ or $x \in \Bbb{R} \setminus \overline{S}$, as $\Bbb{R} = \overline{S} \cup \left(\Bbb{R} \setminus \overline{S}\right)$. If $x \in \Bbb{R} \setminus \overline{S}$ then $x \in \text{ext}(S)$ by definition. Else, $x \in \Bbb{R}\setminus \text{ext}(S)$ Can you argue from here that either $x \in \Bbb{R} \setminus \overline{\text{ext}(S)}$ or $x \in \overline{S} \setminus \text{int}(S)$? It may be helpful to note that  $$\Bbb{R} \setminus \overline{\text{ext}(S)} \subseteq \Bbb{R} \setminus \text{ext}(S)$$
A: Let $S$ be any subset of $\mathbb{R}$. If $x \in \mathbb{R}$ is any point, then one of three mutually exclusive things can happen:


*

*There is some $r > 0$ such that $(x - r, x +r) \subseteq S$. Then by definition, $x \in \operatorname{int}(S)$.

*There is some $r > 0$ such that $(x -r, x + r) \subseteq \mathbb{R} \setminus S$. Then $x \in \operatorname{int}(\mathbb{R}\setminus S) = \operatorname{ext}(S)$.

*For all $r>0$, $(x-r, x+r) \nsubseteq S$ and $(x-r, x+r) \nsubseteq \mathbb{R}\setminus S$, or equivalently, for all $r>0$, $(x-r,x+r) \cap S \neq \emptyset$ and $(x-r,x+r) \cap \mathbb{R}\setminus S \neq \emptyset$. This by definition means $x \in \partial S$.
So $x$ is always in one of the three sets, and these sets are disjoint. Of course, any of them can be empty, or even two of them (for $S = \mathbb{Q}$ the interior and exterior are empty, and all points of $\mathbb{R}$ are boundary points). 
