I'm trying to show that a point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement.
Let $\overline{E}$ denote the closure of $E$ and $E^\circ$ be the interior of $E$. Then the boundary is defined to be $\overline{E} \setminus E^\circ$.
So what I need to show is $\overline{E} \setminus E^\circ = \overline{E} \cap \overline{(S \setminus E)}$.
After looking at this link: A point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement. I am confused about how the OP concluded that he/she needed to show $S \setminus E^\circ = \overline{(S \setminus E)}$.
Could someone explain why?
Best Answer
$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Suppose that you can show that $S\setminus\int E=\cl(S\setminus E)$. Then
$$(\cl E)\setminus\int E=\big((\cl E)\cap S\big)\setminus\int E=(\cl E)\cap(S\setminus\int E)=(\cl E)\cap\cl(S\setminus E)\;,$$
which is exactly what you want.
To see where the idea comes from, note that both $(\cl E)\setminus\int E$ and $(\cl E)\cap\cl(S\setminus E)$ are subsets of $\cl E$. If $A$ is any set disjoint from $\cl E$, therefore,
$$(\cl E)\setminus\int E=(\cl E)\cap\cl(S\setminus E)\quad\text{iff}\quad A\cup\Big((\cl E)\setminus\int E\Big)=A\cup\Big((\cl E)\cap\cl(S\setminus E)\Big)\;,$$
and a clever choice of $A$ might give us sets that are easier to think about than $(\cl E)\setminus\int E$ and $(\cl E)\cap\cl(S\setminus E)$. In particular, if we let $A=S\setminus\cl E$, we can compare the set,
$$(S\setminus\cl E)\cup\Big((\cl E)\setminus\int E\Big)\;,$$
which is simply all of $S$ except $\int E$, or $S\setminus\int E$, with the set
$$(S\setminus\cl E)\cup\Big((\cl E)\cap\cl(S\setminus E)\Big)\;,$$
which is simply $\cl(S\setminus E)$, and try to show that these are equal.