# 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

• Denying that a set is not closed does not imply it is open. Commented Feb 12, 2016 at 19:36
• What does that even mean? If you're not open, then how are you not closed? Commented Feb 12, 2016 at 19:39
• The interval $(1,3]$ is neither open nor closed.
– J126
Commented Feb 12, 2016 at 19:41
• Okay.. so is the only way to prove the claim by doing it constructively? Showing that $\partial A$ contains all limit points of $A$? Commented Feb 12, 2016 at 19:42
• You could also try to show that the complement is open. How do you define the boundary of a set?
– Josh
Commented Feb 12, 2016 at 19:43

## 2 Answers

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 $$M = (\mathrm {int} A) \cup (\mathrm{int} \,M-A) \cup \partial A$$

• Is M taken to be the universal set? Commented Feb 12, 2016 at 19:54
• Yes, I believe you are working with metric space, taking $A \subseteq M$ and $M$ being the metric space in question. Commented Feb 12, 2016 at 19:56
• I don't believe we have explicitly mentioned metric spaces yet in my course. We have been working with the standard euclidean metric and norm thus far (I think). Commented Feb 12, 2016 at 19:57
• If you're working with a normed space, take $d(x,y) = \| x - y \|$ and this will define a metric. Commented Feb 12, 2016 at 20:01
• That's correct. It's not difficult to show this and the union of open sets is open. Then the complement of an open set is closed, and you're done. Commented Feb 14, 2016 at 12:33

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.