Prove that, if {$x_n$}, {$y_n$}, and {$z_n$} are sequences such that $y_n = x_{2n}$ and $z_n = x_{2n-1}$ for $n \in \mathbb{N}$, and $y_n \rightarrow a$ and $z_n \rightarrow a$, then $x_n \rightarrow a$.

My gut says squeeze theorem, but I cannot seem to demonstrate that one sequence is greater and one is less than?

Perhaps this is not the approach, is there a better way?


  • $\begingroup$ Note that all terms get near $a$ as $a \to \infty$, both those with even index and with odd index. $\endgroup$ – coffeemath Feb 28 '13 at 16:14
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    $\begingroup$ @coffeemath. You probably meant to say: as $n\to\infty$. $\endgroup$ – T. Eskin Feb 28 '13 at 16:22
  • $\begingroup$ is this logic enough for a formal proof? $\endgroup$ – Peej Gerard Feb 28 '13 at 16:26
  • $\begingroup$ Squeeze theorem might make sense when you have a "termwise" ordering of the sequences, this is, something of the form $a_n\leq b_n\leq c_n$. If $a_n$ and $c_n$ converge to $a$ then $b_n$ would too. With the hypothesis you specified, you can't deduce any such ordering. $\endgroup$ – fidbc Feb 28 '13 at 16:44

Following @coffeemath's suggestion. Fix $\varepsilon>0$, since $\{y_n\}$ and $\{z_n\}$ converge to $a$ you know there are $N_y$ and $N_z$ st $|y_{n}-a|<\varepsilon$ for all $n\geq N_y$ and $|z_{n}-a|<\varepsilon$ for all $n\geq N_z$.

Now, since

$$x_n=\begin{cases}y_k&& \text{if }n=2k\\ z_k&&\text{if }n=2k-1\end{cases}$$

Then let $N_x=2\max\{N_y,N_z\}$. This would imply $|x_n-a|<\varepsilon$ for all $n\geq N_x$.

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Here is a different approach. By contradiction.

Assume $\{x_n\}\not\rightarrow a$. Therefore there is a subsequence of $\{x_n\}$ that does not converge to $a$. This subsequence must contain an infinite number of even (or odd) indexed terms from $\{x_n\}$ and would therefore contain a subsequence of $\{y_n\}$ (or $\{z_n\}$). This would imply that $\{y_n\}$ (or $\{z_n\}$) does not converge to $a$, a contradiction.

Thanks to @N.S. for pointing out an error in the first version.

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  • $\begingroup$ The part "would therefore be a subsequence" it is actually incorrect. What you probably mean is "would therefore contain a subsequence" $\endgroup$ – N. S. Feb 28 '13 at 16:44
  • $\begingroup$ Well, if all the terms in such subsequence are even indexed terms, say $\{x_{2k_i}\}$, then it is a subsequence of $\{y_n\}$, namely $\{y_{k_i}\}$. Similarly if all the terms are odd indexed. I think I do mean "be a subsequence", don't you think? $\endgroup$ – fidbc Feb 28 '13 at 16:48
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    $\begingroup$ $x_{3n}$ contains infinitely odd and even indexed terms but is neither a subsequence of $x_{2n}$ nor a subsequence of $x_{2n+1}$..... More exactly, what if some indexes are odd and some are even? $\endgroup$ – N. S. Mar 1 '13 at 0:35
  • $\begingroup$ That is true, I'll fix the answer. Thanks for pointing that out! $\endgroup$ – fidbc Mar 1 '13 at 1:48

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