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?
In order for this to be true we should have that if then there are disjoint open balls one containing p and the other containing q.
If is a limit point of and there is an open ball then by definition there is such that .
Now there is an open ball such that .
Now you should be able to complete the proof.
Recall the definition of limit point of the set .
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
The question is about the derived set, , the set of all accumulation points of .
It is not about boundary points.
The problem is to show that is closed in a metric space.
Your proof does not do that. It leaves out any accumulation point in .
P.S BTW. Not every boundary point is an accumulation point.
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.