# Prove that the sequence $\{ z_n\}_{n\in \mathbb{N}}$ is cauchy?

I am given the following problem

Let (X; d) be a metric space, and let $\{ y_n\}_{n\in \mathbb{N}}$ and $\{ x_n\}_{n\in \mathbb{N}}$ be two sequences in $(X, d)$, both converging towards $a\in X$. Let $\{ z_n\}_{n\in \mathbb{N}}$ be the sequence defined by $$z_n = \left\{\begin{array}{lc} x_{(n+1)/2} & n \ \text{is odd} \\ y_{n/2} & n \ \text{is even}\end{array} \right..$$ Prove that $\{ z_n\}_{n\in \mathbb{N}}$ is Cauchy in $(X,d)$.

I have never quite understood how to prove that sequences is Cauchy, even the definition is a tad blurry for me. I know I need to show that there exists an $\epsilon>0$ such that $n,m \Rightarrow N$ implies $x(n,m)<\epsilon$ for every $N$. But I have problems getting started, every hint or nudge is welcome =)

Sorry for posting homework questions here, but when I do not understand something I want to learn it.

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Try to prove that $z_n$ converges to $a$. – Chris Eagle Oct 10 '12 at 9:44
Either $z_n$ is odd or even. If $z_n$ is even we can set $n=2k$, for some $k\in\mathbb{N}$. Then $z_{2k}=y_k$. We know that $y_k$ converges towards $a$, for every value of k and hence $z_n$ converges for every even integer. Similarly we obtain that $z_{2k-1}=x_n$, and the same argument follows. Therefore $z_n$, converges towards $a$ for every value of $n$. Is this enough? Seems very shallow. – N3buchadnezzar Oct 10 '12 at 9:55

Take any $\varepsilon>0$. Since $\{x_n\}_{n\in\mathbb{N}}$ and $\{y_n\}_{n\in\mathbb{N}}$ are both converging to $a$, there exists $N_\varepsilon\in\mathbb{N}$ such that for any $m>N_\varepsilon$, both $|y_m-a|$ and $|x_m-a|$ are less than $\varepsilon$. In particular:

$$|x_m-y_m| = |(x_m-a)-(y_m-a)| \leq |x_m-a|+|y_m-a| < 2\varepsilon.$$

Suppose now $b>a>2N_\varepsilon$. If $a$ and $b$ have the same parity, clearly:

$$|z_b-z_a| < 2\epsilon.$$

If $b$ is even (say $b=2k$) and $a$ is odd (say $a=2j-1$), we have:

$$|z_b-z_a| = |y_k - x_j| \leq |y_k - y_j|+|y_j-x_j| < 4\epsilon,$$

and the same holds if $b$ is odd and $a$ is even.

In conclusion, for any $\epsilon>0$ there exists $N_\varepsilon$ such that for any $a,b>2N_\epsilon$

$$|z_b-z_a|<4\varepsilon$$

holds, so $\{z_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence.

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Say, for all $n > N$, $d(x_n, a) < \epsilon$ and $d(y_n, a) < \epsilon$. What does that tell you about $d(z_m, a)$ for any $m > 2N$?

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