# Homework Help: Set of all the limit points of Set E. Prove that its closed.

1. Aug 30, 2010

### michonamona

Is it correct to make the following statement?

If a point x in E is not a limit point of E, then any neighborhood V of x will--at most--contain finitely many points of E.

Thus, its possible for V to contain only one point, namely, x.

Thanks,

M

2. Aug 30, 2010

### CompuChip

The usual definition of a limit point is as follows
Negating that statement: a point x in X is not a limit point of S, if there exists an open set containing x, but no other points of S.

If you think about a metric space (as I like to do, because I have good intuition for those), this is like saying that x is some finite distance away from all other points, so you will never find a Cauchy sequence converging to it.

I think your statement is not necessarily true, as you can take V = E (let the neighbourhood be all of the space) and E in general contains infinitely many points.*

* I was wondering if $X = (0, 1) \cup \{ 2 \}$ is a counterexample... at first I thought X is not a neighbourhood of 2, but then I was wondering if {2} is not an open set in the inherited topology (it is (1.5, 2.5) intersected with X, for example) which would make X a neighbourhood... it's been a long time since I did topology, so I'm in doubt now

Last edited: Aug 30, 2010
3. Aug 30, 2010

### HallsofIvy

You can actually say a bit more than that- there exist (not just "it is possible") a neighborhood, V, of x that contains only the single point x of E.

Take any neighborhood, U, of x. Since it contains only a finite number of points of E in U, there exist a point, y, in E and U, other than x, such that d(x, y) is minimum. Take V to be the ball about x with radius less than d(x,y).