# Convergent sequence and limit points

bedi

## Homework Statement

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

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## The Attempt at a Solution

We want to show that every neighbourhood around p contains infinitely many $p_{n}$'s.

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

Is this valid?

Staff Emeritus
Homework Helper

## Homework Statement

Prove that if $E \subset X$ and if p is a limit point of E, then there is a sequence $\{p_{n}\}$ in $E$ such that $p=\lim_{n\to\infty}\{p_{n}\}$

Notation issue: when writing $p=\lim_n p_n$, we don't write brackets around the $p_n$

(I presume that there is an invisible "$p_{n} \rightarrow p$ 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 $p_{n}$'s.

By the definition of convergence we can choose infinitely many N's so that infinitely many $p_{n}$'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 $(p_n)_n$. You act like the $p_n$ are given, but they are not. You must show that such a sequence actually exist.

bedi
What if the actual question were p_n converges to p implies that p is limit point of range of {p_n}?

Staff Emeritus
That is not even true. Take $p_n=2$ for all n. Then $p_n\rightarrow 2$, but 2 is not a limit point of $\{p_n~\vert~n\}$.
What is true is: if $p_n\rightarrow p$ and if $p_n\neq p$ for all n, then p is a limit point of $\{p_n~\vert~n\}$.