# Let $\{x_n\}$ be a sequence in a metric space such that $\{x_{2n}\}$, $\{x_{2n+1}\}$ and $\{x_{3n}\}$ all convergent. Show that $\{x_{n}\}$ converges

I don't know how to begin this. My thoughts are that since there are three convergent subsequences, we could use some knowledge about their limits to construct our convergent sequence but I'm not sure how to prove $x_n$ also converges. Is there a property of metric spaces I should use, or something about Bolzano-Weierstrass theorem (for metric spaces)?

EDIT: As noted in a comment I am not allowed to assume they all converge to the same limit.

• You can't assume that all the subsequences converge to the same limit $L$. – sharding4 Mar 7 '18 at 2:16
• True.. Is there something to notice about the $2_n$, $2_{n+1}$,$3_n$ that I should take advantage of? – Jokus Mar 7 '18 at 2:17
• You can't assume, but you can prove it. Assume that "even" and "odd" subsequences have different limits, and use convergence of the third subsequence to arrive to a contradiction. – user58697 Mar 7 '18 at 2:24

Assume that $x_{3n}\rightarrow L$ and $x_{2n}\rightarrow M$, $x_{2n+1}\rightarrow N$, then the sequence $\{x_{6n}\}$ is a convergent subsequence of both $\{x_{3n}\}$ and $\{x_{2n}\}$, so $L=M$. The sequence $\{x_{6n+3}\}$ is a convergent subsequence of both $\{x_{3n}\}$ and $\{x_{2n+1}\}$, so $L=N$, so $L=M=N$.
• $\{x_{3n}\}$ isn't actually a subsequence of either of those, is it? – sharding4 Mar 7 '18 at 2:26
• Split again to get $\{x_{6n}\}$ and $\{x_{6n+3}\}$ then it goes through, thanks for reminding. – user284331 Mar 7 '18 at 2:31