1. ## Prove the following

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

2. Originally Posted by Jdg6057
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 $x\in\mathbb{R}$ is a boundary point of $S$ iff every neighborhood around $x$ contains points in $S$ and $R\backslash S$. A point $x\in\mathbb{R}$ is a boundary point of $\mathbb{R}\backslash S$ iff every neighborhood around $x$ contains points in $\mathbb{R}\backslash S$ and $S$. What can you conclude from this?

b) A point $x\in S$ is an interior point of $S$ iff there exists a neighborhood around $x$ that is a proper subset of $S$. Compare this definition with the above definition and deduce that $int(S)\cap bd(S)=\emptyset$.

c) $\Rightarrow$ If $S$ is open, then every point in $S$ is an interior point, so $bd(S)=\emptyset$. $\emptyset\subset\mathbb{R}\backslash S$ is trivially true.

$\Leftarrow$ If $bd(S)\subset\mathbb{R}\backslash S$, then $bd(S)\cap S=\emptyset$, meaning that every point in $S$ is an interior point, so $S$ is open.