# About sequences in metric spaces

Let $(x_{n})_{n\in\mathbb{N}}$ sequence in a metric space $(X,d)$ such that the subsequences $(x_{2n})_{n\in\mathbb{N}}$, $(x_{2n+1})_{n\in\mathbb{N}}$ and $(x_{3n})_{n\in\mathbb{N}}$ are convergents. Prove that $(x_{n})_{n\in\mathbb{N}}$ is convergent.

-
Do you have any ideas? – JHance Aug 11 '14 at 2:43

$(x_{6n})$ is a subsequence of both $(x_{2n})$ and $(x_{3n})$, which must be convergent, so $(x_{2n})$ and $(x_{3n})$ converge to the same value.

Similarly, $(x_{6n+3})$ is a subsequence of both $(x_{2n+1})$ and $(x_{3n})$, which must be convergent, so $(x_{2n+1})$ and $(x_{3n})$ converge to the same value.

Thus we have: $(x_{2n})$ and $(x_{2n+1})$ converge to the same value. Now it is easy to see that $(x_n)$ converges to the same value with $(x_{2n})$ and $(x_{2n+1})$.

-

We have that $$x_{2n}\to c,\qquad x_{2n+1}\to d.$$ Since $x_{3n}\to e$, $x_{6n}\to e$, hence $e=c$.

Since $x_{3n}\to e$, $x_{6n+3}\to e$, hence $d=e$ and $c=d$, giving $x_n\to c=d=e.$

-

Assume that $x_{2n}\to l$ then $x_{6n}\to l$ which implies that $x_{3n}\to l$ (since $\{x_{3n}\}$ converges). Now we have $x_{6n+3}=x_{3(2n+1)}\to l$ and $x_{6n+3}=x_{2(3n+1)+1}\to \lim_{n\to \infty} x_{2n+1}$.

Now we know that $\{x_{2n}\}$ and $\{x_{2n+1}\}$ which means that $\{x_n\}$ converges.

-