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Suppose that you have a countable family of sequences each one of these converges to the same point: $$ x_n^{(i)} \rightarrow x \hspace{1cm} \forall i\in\mathbb{N}. $$ The sequence $\{x_n^{(n)}\}$ is such that $x_n^{(n)} \rightarrow x$? (i.e. the $n^{th}$ element of the new sequence is the $n^{th}$ element of the $n^{th}$ sequence.) Why?

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up vote 5 down vote accepted

No. For example, consider $\{x^{(i)}_n\}_{n=1}^\infty =\{ 0,0,\ldots,0,\underbrace{1}_{i^{\rm th}\text{-term}},0,0,\ldots\}$.

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Or even worse, $\{0,0,\ldots,0,i,0,0,\ldots\}$. – MJD Oct 8 '12 at 19:15
That even worse statement is shocking. – Rojas Azules Oct 8 '12 at 19:23

Let $x_n^{(i)}=\frac i n c_i$ Then for fixed $i$ this goes to $0$ as $n\to\infty$, but $x_n^{(n)}=c_n$ is an arbitrary sequence.

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Yep. +1. $ $ $ $ – Did Oct 8 '12 at 19:03

Let $X$ be any topological space with more than one point, let $x=\langle x_k:k\in\Bbb N\rangle$ be a convergent sequence in $X$ with limit $p$, and let $q\in X\setminus\{p\}$. For $n\in\Bbb N$ let $x^{(n)}=\left\langle x^{(n)}_k:k\in\Bbb N\right\rangle$, where $$x^{(n)}_k=\begin{cases}q,&\text{if }k\le n\\x_{k-n-1},&\text{otherwise}\;.\end{cases}$$ Clearly $x^{(n)}\to p$ for each $n\in\Bbb N$, but $x^{(n)}_n=q$ for each $n\in\Bbb N$, so $\left\langle x^{(n)}_n:n\in\Bbb N\right\rangle\to q$. Change the definition so that

$$x^{(n)}_k=\begin{cases}q,&\text{if }k\le n\text{ and }n\text{ is even}\\p,&\text{if }k\le n\text{ and }n\text{ is odd}\\x_{k-n-1},&\text{otherwise}\;,\end{cases}$$

and $\left\langle x^{(n)}_n:n\in\Bbb N\right\rangle$ doesn’t converge at all: its terms are alternately $q$ and $p$.

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