Show that $\partial A$ is always a closed set

Question

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

What does that even mean? If you're not open, then how are you not closed? — Javier, Feb 12 '16 at 19:39
Okay.. so is the only way to prove the claim by doing it constructively? Showing that $\partial A$ contains all limit points of $A$? — Javier, Feb 12 '16 at 19:42
You could also try to show that the complement is open. How do you define the boundary of a set? — Josh, Feb 12 '16 at 19:43
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 $A^c$ — Javier, Feb 12 '16 at 19:54
"If x is not open then x is closed" is true if x is a door. Closed sets are the complements of open sets.In general there are sets which are neither open nor closed .In the reals [0,1) is neither open nor closed. — DanielWainfleet, Feb 12 '16 at 20:42

Aaron Maroja · Accepted Answer · 2016-02-12T19:55:03.240

2

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

edited Feb 12 '16 at 19:55

answered Feb 12 '16 at 19:52

Aaron Maroja

17,571

Is M taken to be the universal set? – Javier Feb 12 '16 at 19:54
Yes, I believe you are working with metric space, taking $A \subseteq M$ and $M$ being the metric space in question. – Aaron Maroja Feb 12 '16 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). – Javier Feb 12 '16 at 19:57
If you're working with a normed space, take $d(x,y) = | x - y |$ and this will define a metric. – Aaron Maroja Feb 12 '16 at 20:01
...which in turn defines a topology, and this is all that is of interest here. – Alex Provost Feb 12 '16 at 20:12
Can I prove the claim at hand by showing both int(A) and (intM - A) are both open? – Javier Feb 14 '16 at 03:00
1

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. – Aaron Maroja Feb 14 '16 at 12:33

score 0 · Answer 2 · answered Feb 12 '16 at 20:17

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.

Show that $\partial A$ is always a closed set

2 Answers2