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

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The discussion centers on the characterization of limit points in topology, specifically addressing the statement that if a point x in a set E is not a limit point, then any neighborhood V of x contains at most finitely many points of E. The participants clarify that it is indeed possible to construct a neighborhood V that contains only the point x itself, reinforcing the definition of limit points. The conversation also touches on the implications of metric spaces and the construction of neighborhoods in relation to limit points.

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michonamona
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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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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 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
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Last edited:
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).
 

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