Prove that a closed set contains all of its boundary points while an open set contains none of it's boundary points.
So the book gives me a definition of a boundary point as: Let $A$ be a subset of $\mathbb{R}$. A point $x_0$ is called a boundary point of a set $A$ if every ball with center at $x_0$ contains points of $A$ and points of the compliment of $A$.
So here's my attempt:
Suppose a closed set A contains all of it's boundary points but the point k which is not included in the set. Then k is in the set which compliments A. So k is some ball that is centered at k and is in the compliment of A. (here is where I was stuck.)
Suppose A is an open set. (Here is where I was stuck .)
Both proofs are contradictions so…
Best Answer
First assume $A$ is closed. Assume, for a contradiction, that there is $x$ in the boundary of $A$ such that $x \notin A$. Then $x \in A^c$ which is open so there exists some ball centred at $x$ that is entirely in $A^c$. Then this ball contains no points in $A$ contradicting the definition of the boundary.
Now assume $A$ is open. $\forall x \in A$ there is $\epsilon > 0$ such that the ball of radius $\epsilon$ is contained in $A$. But then that ball contains no points in $A^c$ so $x$ cannot be in the boundary of $A$.