Proving that $y_n$ converges given $y_{2n}, y_{2n+1}$ converges

Question

Suppose we have $\mathrm{lim}_{n \rightarrow \infty} y_{2n} = \mathrm{lim}_{n \rightarrow \infty} y_{2n+1} = M \in \mathbb R$. I'm trying to prove from this that $\mathrm{lim}_{n \rightarrow \infty} y_{n} = M$. This is my attempt:

Fix $\varepsilon > 0$. Let $m_1,m_2\in \mathbb N$ be such that for $n\geq m_1$, $|x_{2n}-M|<\varepsilon$, and for $n\geq m_2$, $|x_{2n+1}-M|<\varepsilon$. Taking $m_0=\max(2m_1,2m_2+1)$, it follows that for $n\geq m_0$, $|x_n-M|<\varepsilon$, and so $x_n\rightarrow M$ as $n\rightarrow\infty$.

The step I'm uncertain of is $m_0=\max(2m_1,2m_2+1)$. I know that I have to set $m_0$ s.t. I am covering all even naturals after $m_1$ and all odd naturals after $m_2$ however I am not sure if this choice of $m_0$ satisfies both these requirement. Intuitively it seems quite obvious but rigorously I've not been able to show it.

Answer 1

The index $m_0$ you defined works. Let $m\geq m_0$. If $m$ is even then you can write $m=2n$. Since $m\geq 2m_1$ we have $n\geq m_1$ and hence $|x_m-M|=|x_{2n}-M|<\epsilon$. If $m$ is odd then you can write $m=2n+1$. Since $m\geq 2m_2+1$ we have $n\geq m_2$ and hence $|x_m-M|=|x_{2n+1}-M|<\epsilon$. So anyway $|x_m-M|<\epsilon$.

Answer 2

Your answer if perfectly correct!! Also you can prove like you did that if $m$ subsequences of $\{x_n\}$ converges to the same point $M$ and the union of all the index of the subsequences cover all $\mathbb{N}$ then $\{ x_n \} \rightarrow M$