# Thread: Proof dealing with a closed set

1. ## Proof dealing with a closed set

Show that the set E' of points of accumulation of any set E must be closed.

Not too sure what to do with this; I understand why it is but really don't have a clue how to show it. Help please?

2. What is the definition? What happens if you let x be an element of the boundry? for all epsilon>0. we have that the intersection between that neighbourhood and the set E is nonempty. So...

3. What sort of space are we working with?
There are topological spaces in which this problem is a false statement.

4. Plato, I'm not quite sure how to answer your question. I am not in a typology class, this is for my real analysis class. This problem is just under a section dealing with open and closed sets.

5. I think that he's just talking about subsets of the real numbers.

6. Originally Posted by steph3824
for my real analysis class. This problem is just under a section dealing with open and closed sets.
A real analysis class would at the very least require a metric space.
In order for this to be true we should have that if $\displaystyle p\not= q$ then there are disjoint open balls one containing p and the other containing q.

If $\displaystyle p$ is a limit point of $\displaystyle E'$ and there is an open ball $\displaystyle p\in \mathcal{B}$ then by definition there is $\displaystyle q\in \mathcal{B}\cap E'$ such that $\displaystyle q\not= p$.
Now there is an open ball $\displaystyle \mathcal{C}$ such that $\displaystyle q\in \mathcal{C}\subset \mathcal{B}\setminus\{p\}$.

Now you should be able to complete the proof.
Recall the definition of limit point of the set $\displaystyle E$.

7. I'm not really an analysis guy, so this could be wrong...

But can't you just write $\displaystyle E^{\prime}=\overline{E}\cap\overline{X\setminus E}$?

8. Simplified this and redid it below with original post. Didn't know how to delete this.

9. Originally Posted by steph3824
Show that the set E' of points of accumulation of any set E must be closed.

Not too sure what to do with this; I understand why it is but really don't have a clue how to show it. Help please?
$\displaystyle E^'$ = $\displaystyle E_0^c \cap \overline{E}$

$\displaystyle E_0^c$ is the complement of the interior (open) and so is closed

$\displaystyle \overline{E}$ is the closure of E and so is closed.

Intersection of closed sets is closed.

10. Originally Posted by Hartlw

$\displaystyle E^'$ = $\displaystyle E_0^c \cap \overline{E}$ This is not correct.
$\displaystyle E_0^c$ is the complement of the interior (open) and so is closed
$\displaystyle \overline{E}$ is the closure of E and so is closed. Intersection of closed sets is closed.
If $\displaystyle E=[0,1)$ then $\displaystyle E_0^c \cap \overline{E}=\{0,1\}$ the boundary.
BUT $\displaystyle E'=[0,1].$

11. Originally Posted by Plato
If $\displaystyle E=[0,1)$ then $\displaystyle E_0^c \cap \overline{E}=\{0,1\}$ the boundary.
BUT $\displaystyle E'=[0,1].$
The complement of the interior is all points except the interior.
The closure is the interior E0 plus the boundary E'. The intersection is the boundary.

if E=[0,1), then E' = 0 and 1

E' is the set of all accumulation points of E, which is the same for (0,1) or [0,1) or [0,1]: 0 and 1

12. Originally Posted by Hartlw
The complement of the interior is all points except the interior. The closure is the interior plus the boundary. The intersection is the boundary.
All of that is very true. But it has nothing to do with this OP.
The question is about the derived set, $\displaystyle A'$, the set of all accumulation points of $\displaystyle A$.
It is not about boundary points.

The problem is to show that $\displaystyle A'$ is closed in a metric space.
Your proof does not do that. It leaves out any accumulation point in $\displaystyle A^o$.

P.S BTW. Not every boundary point is an accumulation point.

13. E' is the set of all accumulation points of E, which by definition is the boundary.

14. Originally Posted by Hartlw
This statement is also incorrect.
E' is the set of all accumulation points of E, which is the same for (0,1) or [0,1) or [0,1]: 0 and 1
If $\displaystyle E=[0,1)\cup \{2,3\}$ then $\displaystyle E'=[0,1]$.

Originally Posted by Hartlw
This statement is also incorrect.
E' is the set of all accumulation points of E, which by definition is the boundary.
I think that you are confused about the definition of accumulation point.

15. Originally Posted by Plato
If $\displaystyle E=[0,1)\cup \{2,3\}$ then $\displaystyle E'=[0,1]$.

I think that you are confused about the definition of accumulation point.
You are correct of course. Interior points are also accumulation points. All boundary points are accumulation points but not the other way around.

The complement of the interior includes the boundary and exterior points, and my original proof was correct.

Unfortunateley I can't copy it and I just don't feel like plowing through the code again, so it will have to remain buried (#9).

My original statement:
"E' is the set of all accumulation points of E, which by definition is the boundary" shoud read:
E' is the set of all accumulation points of E, which by definition includes the boundary.

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