If $\{x_{2n}\}$ , $\{x_{2n+1}\}$ and $\{x_{3n}\}$ converge, does $\{x_n\}$ converge? 
If $\{x_{2n}\}$ , $\{x_{2n+1}\}$ and $\{x_{3n}\}$ converge, does $\{x_n\}$ converge?

Are these sufficient conditions for convergence in any metric space? Thanks.
 A: Yes.

Suppose 
\begin{align*}
&x_{2n}\;\text{converges to}\;a\\[4pt]
&x_{2n+1}\;\text{converges to}\;b\\[4pt]
&x_{3n}\;\text{converges to}\;c\\[4pt]
\end{align*}

Since $(x_{2n})$ and $(x_{3n})$ share a common infinite subsequence, it follows that $a=c$.

Similarly, since $(x_{2n+1})$ and $(x_{3n})$ share a common infinite subsequence, it follows that $b=c$.

Thus, $a = b = c$.

Since $x_{2n}$ converges to $c$, and $x_{2n+1}$ converges to $c$, it follows that $x_n$ converges to $c$.
A: Let $a=\lim x_{2n}, b=\lim x_{2n+1}$ and $c=\lim x_{3n}$.
$(x_{6n})$ is a subsequence of $(x_{2n})$ and of $(x_{3n})$. Hence: $a=c$
$(x_{6n+3})$ is a subsequence of $(x_{2n+1})$ and of $(x_{3n})$. Hence: $b=c$.
Its now your turn to show that $(x_n)$ has the limit $a$
A: Yep. Let
$$\lim_{n\to\infty} x_{2n} = a_1,$$
$$\lim_{n\to\infty} x_{2n+1} = a_2,$$
$$\lim_{n\to\infty} x_{3n} = a_3$$
Now, $\{x_{6n}\}$ is a subsequence of both $\{x_{2n}\}$ and $\{x_{3n}\}$. So $a_1=a_3$. And $\{x_{6n+3}\}$ is a subsequence of both $\{x_{2n+1}\}$ and $\{x_{3n}\}$. So $a_2=a_3=a_1.$ Let $a=a_1=a_2=a_3.$ Then
$$\lim_{n\to\infty} x_{n} = a.$$
because
$$\lim_{n\to\infty} x_{2n} = a = \lim_{n\to\infty} x_{2n+1}$$
and as $n \in \mathbb{N}$ is either even or odd, $n=2k$ or $n=2k+1$ so
$$\lim_{n\to\infty} x_{n} = 
\left\{\begin{array}{lr} 
\lim_{k\to\infty} x_{2k} & \text{n even}\\
\lim_{k\to\infty} x_{2k+1} & \text{n odd} \end{array}\right\} = a.$$
