# Two convergent sequences in a metric space.

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.

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