Show that $\partial A$ is always a closed set First, I believe there are at least two ways to prove this result. One, constructively, by showing that $\partial A$ contains all limit points. The other, by contradiction, is to suppose that $\partial A$ is open. I chose this direction because it seemed easier to me.
So, I have: Suppose $\partial A$ is open. That is, $\forall x \in \partial A$, $\exists \epsilon > 0$, s.t: $B_\epsilon (x) \subset \partial A$. Now, to show the contradiction, I want to show that if $y \in B_\epsilon (x)$, then $y \notin \partial A$. This is where I am having trouble. For instance, I know that if $y \in B_\epsilon (x)$, then $\parallel x - y \parallel < \epsilon$. I want to use the definition of $\partial A$ := {$x \in R^d | \forall \epsilon > 0, B_\epsilon (x)$ contains points in $A$ and $A^c$}, but I can't see where to go from here. Any help is appreciated. Thanks.
Edit: Attempt 2:
Let $x \in R^d$. Consider $B_{1/n} (x) \in \partial A$. See that $B_{1/n} (x) \longrightarrow x$ as $n \longrightarrow \infty$ $\Longrightarrow$ $x \in \partial A$. Thus, $\partial A$ contains all limit points and so it is a closed set.
I feel like this may not be right but I think I am close, any input is appreciated.Pls help so I don't fail this class and have to repeat for a third time
 A: As mentioned "closed" is not the opposite of "open". Take for instance $[a,b) \subset \mathbb R$. Is it closed?
Hint: In order to show that $\partial A$ is closed notice that the complement of $\partial A$ is $(\mathrm {int} A) \cup (\mathrm {int} \,M - A)$. Since we may write $M$ (space in question) as the disjoint union $$M  = (\mathrm {int} A) \cup (\mathrm{int}  \,M-A) \cup \partial A$$
A: Using your definition of boundary, 
"I am defining the boundary of the set as vectors in $R^d$ s.t: for all positive epsilon, the open ball around a point contained within the boundary contains points from $A$ and $A^c$",
we will show that it contains its limit points. 
If $p$ is a limit point of the boundary then for any $\epsilon\gt 0$ the open ball of radius $\epsilon$ around $p$ contains a point of the boundary, call it $x$. Fix $\epsilon$ and choose $\delta\gt 0$ such that the open ball of radius $\delta$ around $x$ is completely contained in the open ball of radius $\epsilon$ around $p$.Since $x$ is in the boundary, the ball of radius $\delta$ contains points from both $A$ and $A^c$. But this ball is contained in the ball of radius $\epsilon$ around $p$ so by your definition, $p$ is in the boundary.
Therefore, the boundary is closed.
