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Convergent sequence and limit points

  • Thread starter bedi
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Homework Statement



Prove that if [itex]E \subset X[/itex] and if p is a limit point of E, then there is a sequence [itex]\{p_{n}\}[/itex] in [itex]E[/itex] such that [itex]p=\lim_{n\to\infty}\{p_{n}\}[/itex] (I presume that there is an invisible "[itex]p_{n} \rightarrow p[/itex] implies that" at the beginning of the sentence).

Homework Equations



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The Attempt at a Solution



We want to show that every neighbourhood around p contains infinitely many [itex]p_{n}[/itex]'s.

By the definition of convergence we can choose infinitely many N's so that infinitely many [itex]p_{n}[/itex]'s can be contained in a neighbourhood of p with a correspondent radius.

Is this valid?
 

Answers and Replies

  • #2
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Homework Statement



Prove that if [itex]E \subset X[/itex] and if p is a limit point of E, then there is a sequence [itex]\{p_{n}\}[/itex] in [itex]E[/itex] such that [itex]p=\lim_{n\to\infty}\{p_{n}\}[/itex]
Notation issue: when writing [itex]p=\lim_n p_n[/itex], we don't write brackets around the [itex]p_n[/itex]

(I presume that there is an invisible "[itex]p_{n} \rightarrow p[/itex] implies that" at the beginning of the sentence).
Nope, the sentence is fine how it is. Why do you think that should be added?

The Attempt at a Solution



We want to show that every neighbourhood around p contains infinitely many [itex]p_{n}[/itex]'s.

By the definition of convergence we can choose infinitely many N's so that infinitely many [itex]p_{n}[/itex]'s can be contained in a neighbourhood of p with a correspondent radius.

Is this valid?
No, this is not valid. And I think you're missing the entire point of the exercise. The exercise wants you to actually construct a sequence [itex](p_n)_n[/itex]. You act like the [itex]p_n[/itex] are given, but they are not. You must show that such a sequence actually exist.
 
  • #3
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What if the actual question were p_n converges to p implies that p is limit point of range of {p_n}?
 
  • #4
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3,277
What if the actual question were p_n converges to p implies that p is limit point of range of {p_n}?
That is not even true. Take [itex]p_n=2[/itex] for all n. Then [itex]p_n\rightarrow 2[/itex], but 2 is not a limit point of [itex]\{p_n~\vert~n\}[/itex].

What is true is: if [itex]p_n\rightarrow p[/itex] and if [itex]p_n\neq p[/itex] for all n, then p is a limit point of [itex]\{p_n~\vert~n\}[/itex].
 
  • #5
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Then what about the latter?
 

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