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?
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 $\displaystyle E$.
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
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.
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.