# To show that a sequence $(x_n)$ in a metric space $(X,d)$ is convergent if the three sub-sequences of it are convergent.

Show that a sequence $(x_n)$ in a metric space $(X,d)$ is convergent if the three sub-sequences $\{x_{2n} : n \in \Bbb N \}, \{x_{2n+1} : n \in \Bbb N \}$ and $\{x_{3n} : n \in \Bbb N \}$ of it are convergent.

There is a very well known result in analysis that given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms say $L$ then $\lim_{n\to\infty}a_{n}=L$.

But the above problem does not gives the condition that the odd seq and the even seq converge to the same limit. I am facing difficulty to do the problem. Please Help!!

• Note both the even and odd sequences have subsequences that are subsequences of the "$3n$" sequence. – David Mitra Feb 16 '15 at 13:49
• so both even and odd sequence will converge to the limit which same as that where the seq $x_{3n}$ converges!! – user8795 Feb 16 '15 at 13:55

Since $\{x_{2n}\}$ and $\{x_{2n+1}\}$ are convergent, the subsequence $\{x_{6n}\}$ of $\{x_{2n}\}$ and the subsequence $\{x_{6n+3}\}$ of $\{x_{2n+1}\}$ are convergent. Since $\{x_{6n}\}$ and $\{x_{6n+3}\}$ are subsequences of the convergent sequence $\{x_{3n}\}$, they have the same limit. So $\{x_{2n}\}$ and $\{x_{2n+1}\}$ have the same limit, call it $x$.
Given $\epsilon > 0$, choose $N_1,N_2\in \Bbb Z^+$ such that $d(x_{2n},x) < \epsilon$ for all $n\ge N_1$ and $d(x_{2n+1},x) < \epsilon$ for all $n \ge N_2$. Let $N = \max\{2N_1,2N_2 + 1\}$. If $n \ge N$ and $n$ is even, $n = 2k$, then $k \ge N_1$, and so $d(x_n,x) = d(x_{2k},x) < \epsilon$. If $n \ge N$ and $n$ is odd, $n = 2k + 1$, then $k \ge N_2$, and thus $d(x_n,x) = d(x_{2k+1},x) < \epsilon$. Therefore $d(x_n,x) < \epsilon$ for all $n \ge N$. Since $\epsilon$ was arbitrary, $x_n \to x$.