If S is a subset of the Reals Prove:
a) bd(S) = bd(R\S)
b) int(S) intersect bd(S) = empty set
c) S is open IFF bd(S) is a subset of R\S
a) A point $\displaystyle x\in\mathbb{R}$ is a boundary point of $\displaystyle S$ iff every neighborhood around $\displaystyle x$ contains points in $\displaystyle S$ and $\displaystyle R\backslash S$. A point $\displaystyle x\in\mathbb{R}$ is a boundary point of $\displaystyle \mathbb{R}\backslash S$ iff every neighborhood around $\displaystyle x$ contains points in $\displaystyle \mathbb{R}\backslash S$ and $\displaystyle S$. What can you conclude from this?
b) A point $\displaystyle x\in S$ is an interior point of $\displaystyle S$ iff there exists a neighborhood around $\displaystyle x$ that is a proper subset of $\displaystyle S$. Compare this definition with the above definition and deduce that $\displaystyle int(S)\cap bd(S)=\emptyset$.
c) $\displaystyle \Rightarrow$ If $\displaystyle S$ is open, then every point in $\displaystyle S$ is an interior point, so $\displaystyle bd(S)=\emptyset$. $\displaystyle \emptyset\subset\mathbb{R}\backslash S$ is trivially true.
$\displaystyle \Leftarrow$ If $\displaystyle bd(S)\subset\mathbb{R}\backslash S$, then $\displaystyle bd(S)\cap S=\emptyset$, meaning that every point in $\displaystyle S$ is an interior point, so $\displaystyle S$ is open.