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Show if a sequence converges, then each subsucession converges to the same limit in Euclidean space

Then let $(x_{n}) \in \mathbb{R^n}$ a convergent sequence and $(x_{r}) \in \mathbb{R^n}$ a sub-succession of $(x_{n})$, it should be shown that $(x_{r})$ is convergent and its limit is equal to that of the $(x_{n})$ sequence.

proof:

let $(x_{r}) \in \mathbb{R^n}$ a sub-sequence of $(x_ {n}) \in \mathbb{R^n}$; given an $\epsilon>0$, exists $k \in \mathbb{N}$ such that $(x_{n}) \in B(a,\epsilon)$ for all $k>K$ where $a \in \mathbb{R^n}$ and is the limit of the sequence, that is to say, $\lim x_{k}=a$. now let's see what $\lim x_{r}=a$

I'm stuck, could someone help me?

tqwe
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1 Answers1

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Given $\epsilon>0$, since $(x_n)$ converges to $a$, there is some $k\in\Bbb N$ such that $x_n\in B(a,\epsilon)$ for every $n\ge k$.

Since $(x_r)$ is a subsequence of $(x_n)$, each $r$ in $x_r$ actually is some $n$ in $x_n$ (meaning each $x_r$ is some $x_n$ from the original sequence). Then there is some $l\in\Bbb N$ such that $x_r\in(x_n)_{n\ge k}$, meaning there is a moment when the subsequence has run out of terms $x_n$ with $n<k$. For such terms of $(x_r)$ we have $x_r\in B(a,\epsilon)$, since each $x_r$ with $r\ge l$ represents some $x_n$ with $n\ge k$.

We conclude $(x_r)$ converges to $a$ as well.

Note: This argument is valid for any topological space associated to a metric (or even to a pseudo-metric).

Darsen
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