Proving that $U^o=U \setminus \partial{U}$ (Boundary and interior of a set)

Question

I want to prove that $U^o=U$\ $\partial U$ where $U^o$ is the interior of $U$ and $\partial U$ is the boundary of U where $U$ is a subset of $X$.

I have the definitions $U^o=X \setminus (\overline{X \setminus U})$ and $\partial U=\overline{U} \setminus U^o$

I think that i should use that $A \cap B= A \setminus (X \setminus B)$ and $A \setminus (B \cap C)=(A\setminus B) \cup (A \setminus C)$ where $A, B$ and $C$ are subsets of $X$

So far i have got $U^o=X \setminus (\overline{X \setminus U})=X \setminus (X \setminus U^o $) and don't know where to go from here.

Answer 1

Lemma 1. $U^o\subseteq U$

Proof. I leave as an exercise. $\Box$

Lemma 2. $\overline{U}=U\cup\partial U$

Proof. The "$\supseteq$" inclusion is obvious. For "$\subseteq$" assume that $u\in \overline{U}$ but $u\not\in U$. By lemma 1 $u\not\in U^o$ and thus $u\in\overline{U}\backslash U^o=\partial U$. $\Box$

Lemma 3. $U^o=U\backslash\partial U$

Proof. Lemma 2 implies that $U\backslash\partial U=\overline{U}\backslash\partial U$ and thus $$U\backslash\partial U=\overline{U}\backslash\partial U=\overline{U}\backslash\big(\overline{U}\backslash U^o\big)=U^o$$ The last equality because $U^o\subseteq \overline{U}$. $\Box$