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Question: Let {$x_n$} and {$y_n$} be two convergent sequences in a metric space (E,d). For all $n \in \mathbb{N}$, we defind $z_{2n}=x_n$ and $z_{2n+1}=y_n$. Show that {$z_n$} converges to some $l \in E$ $\longleftrightarrow$ $ \lim_{n \to \infty}x_n$= $\lim_{n \to \infty}y_n$=$l$.

My Work: Since $x_n$ converges, $\exists N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $|x_n-l_1|<\epsilon$. Likewise, since $y_n$ converges, $\exists N_2 \in \mathbb{N}$ s.t. $\forall n\geq N_2$, $|y_n-l_2|<\epsilon$. Because $z_{2n}=x_n$ and $z_{2n+1}=y_n$, then pick $N=\max\{N_1,N_2\}$. Since eventually $2n+1>2n>N$, if $z_n$ converges to $l$, then $|z_n-l|=|x_{n/2}-l|<\epsilon$ because of how we picked our N. Am I correct in this approach and should continue this way or am I wrong? This is a homework problem so please no solutions!!! Any help is appreciated.

My work for the other way: If $\lim_{n \to \infty}x_n=\lim_{n \to \infty}y_n=l$,then we show $\lim_{n \to \infty}$$z_n=l$.Then for $N \in \mathbb{N}$, where $N=max{(\frac{N_1}{2},\frac{N_2-1}{2})}$. Since $|x_n-l|=|y_n-l|=|z_{2n}-l|=|z_{2n+1}-l|<\epsilon$ then for $n\geq N$, $|z_n-l|<\epsilon$.

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4 Answers 4

up vote 2 down vote accepted

You’re working with the right ideas, though what you’ve done so far is insufficient. Remember that you need to prove two things: if $\langle z_n:n\in\Bbb N\rangle\to\ell$, then $\ell_1=\ell_2=\ell$, and if $\ell_1=\ell_2$, then $\langle z_n:n\in\Bbb N\rangle\to\ell_1=\ell_2$. Your argument will be much easier to write and to read if you clearly separate these two parts:

  1. Assume that $\langle z_n:n\in\Bbb N\rangle\to\ell$, and use what you’ve done to show that $\ell_1=\ell_2=\ell$.

  2. Assume that $\ell_1=\ell_2$, and show that $\langle z_n:n\in\Bbb N\rangle$ converges to this number.

You’ve got everything that you need, though it can be improved a bit: you should let $N=\max\{2N_1,2N_2+1\}$, so that taking $n>N$ guarantees that $n/2>N_1$ if $n$ is even and $(n-1)/2>N_2$ if $n$ is odd.

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Hint: for any converging sequence $\,\{a_n\}\,$

$$a_n\xrightarrow [n\to\infty]{} L\Longrightarrow\;\;\; a_{2n}\,,\,a_{2n+1}\xrightarrow [n\to\infty]{} L$$

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You, of course, realise that this is an "if and only if" statement. The work you are showing is only one direction. There is a result that says that if $\{a_n\} \to L$ then any subsequence $\{a_{n_k}\}$ also goes to $L$. It is not hard to prove. Since all the terms after some $N \in \mathbb{N}$ will be in an $\epsilon$-distance from the limit, this will trivially be true for a subset of those terms (namely, the elements of the subsequence). If you prove this or if you are allowed to use this then one implication is trivial if you note that both $\{x_n\}$ and $\{y_n\}$ are subsequences of $\{z_n\}$. Do you have the other implication?

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I edited my original work to add the other way. I wasn't sure If I was on the right track so I wanted to see if I was before adding it. –  Mathstudent Oct 11 '12 at 4:32

Hint Remember the fact that "every convergent sequence is a Cauchy sequence".

$(\Rightarrow)$ Assume $\lim_{n \to \infty} z_n=l$, then notice that

$$ |x_n-l|=|z_{2n}-l|=|(z_{2n}-z_n)+(z_n-l)|\leq |z_{2n}-z_n|+|z_n-l|<\dots $$

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