Set of all the limit points of Set E. Prove that its closed.

In summary, a point x not being a limit point of a subset does not mean that V cannot contain only the single point x of the subset.
  • #1
michonamona
122
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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
 
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  • #2
The usual definition of a limit point is as follows
Wikipedia said:
Let S be a subset of a topological space X. A point x in X is a limit point of S if every open set containing x contains at least one point of S different from x itself.

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 [itex]X = (0, 1) \cup \{ 2 \}[/itex] 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
 
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  • #3
michonamona said:
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
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).
 

FAQ: Set of all the limit points of Set E. Prove that its closed.

What is a limit point?

A limit point is a point in a set that can be approached by infinitely many points in that set. It may or may not be a member of the set itself.

What is a closed set?

A closed set is a set that contains all of its limit points.

How do you prove that a set is closed?

To prove that a set is closed, we need to show that it contains all of its limit points. In other words, every point that can be approached by infinitely many points in the set is also a member of the set.

What is the set of all limit points of Set E?

The set of all limit points of Set E, denoted by E', is the set of all points that can be approached by infinitely many points in Set E.

How do you prove that the set of all limit points of Set E is closed?

We can prove that the set of all limit points of Set E is closed by showing that it contains all of its limit points. This can be done by assuming a point is a limit point of E' and showing that it is also a member of E'.

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