Proving a closed set contains all of it's boundary points? 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...
 A: 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 $A$.
A: In general, if $A$ is a subset of a topological space $X$, the boundary of $A$ is
$$\partial A = \overline A\cap \overline {A^c}, $$
where $\overline A$ denotes the closure of $A$, i.e.
$$\overline A = \bigcap_{\substack{F\supset A\\F \text{ closed} }} F. $$ It is straightforward to show that this definition of boundary is equivalent to the one your book gives. If $A$ is closed, then $A=\overline A$, so $$\partial A\subset \overline A=A.$$ If $A$ is open, then $A^c$ is closed, so $$\partial A\subset \overline{A^c}=A^c.$$
A: Hints for the first part of your question. Assume that $p$ is a boundary point that lies outside $A$. Hence $p \in \complement A$, the complement of $A$. But $A$ is closed, so $\complement A$ is open and therefore $p$ is an interior point of $\complement A$. But any neighborhood of $p$ intersects $A$, so that we reach a contradiction.
