$(n_k)_{k=1}\cup (m_k)_{k=1}=\mathbb{N}$ so $\lim_{n \to \infty}a_n=L\Leftrightarrow \lim_{k \to \infty}a_{n_k}=L \cap \lim_{k \to \infty}a_{m_k}=L$ I need to know if my answer to the question below is correct. Thank for your help.
Question:
Let $(n_k)^{\infty }_{k=1}$ and $(m_k)^{\infty }_{k=1}$ two sequences of indexes stricly increasing and such that: $(n_k)^{\infty }_{k=1}\cup (m_k)^{\infty }_{k=1}=\mathbb{N}$. Prove that: $\lim_{n \to \infty}a_n=L\Leftrightarrow \lim_{k \to \infty}a_{n_k}=L \cap \lim_{k \to \infty}a_{m_k}=L$
My answer:
$(\Rightarrow )$:
We know by theorem that all the subsequences of a sequence $a_n$ s.t. $\lim_{n \to \infty}a_n=L$ converges too toward $L$ so in particular the two subsequences $a_{n_k}$ and $a_{m_k}$
$(\Leftarrow )$:
By contradiction let assume that: $\lim_{n \to \infty}a_n\neq L$. It means by definition:$\exists \epsilon _0 >0 \; s.t. \; \forall N \in \mathbb{N} , \exists n' \in \mathbb{N} \geq N \Rightarrow|a_{n'}-L| \geq \epsilon _0 $ (i)
On the other hand by assumption we know that: $\forall \epsilon >0 \; \exists K \in \mathbb{N} \; s.t. \; \forall \; m_k, n_k > K \Rightarrow |a_{n_k}-L|<\epsilon \; and \; |a_{n_k}-L|<\epsilon$. If it is true for all $\epsilon>0$ it is true in particular for $\epsilon=\epsilon _0$. Now let writte $K_0$ the (or a) $K$ corresponding to the special $\epsilon _0$ that we fixed.(ii)
Now according to (i) it exists at least one $n'>K_0$ s.t. $|a_{n'}-L| \geq \epsilon _0$ . By definition $n' \in \mathbb{N}= \left \{ m_k, n_k>K_0 \cup  m_k, n_k \leq K_0  \right \}$ (it is given $\mathbb{N}= \left \{ m_k, n_k>K_0 \cup  m_k, n_k \leq K_0  \right \}$). Moreover as we fixed in $ \epsilon=\epsilon _0 $ and so on $K_0$ in (ii) this $n'$ can not be one of the indexes $m_k, n_k \leq K_0$. It means that $n' \in \left \{ m_k, n_k>K_0 \right \}$ (or in other word $n' \notin \left \{ m_k, n_k\leq K_0 \right \} $).
!!CONTRADICTION with (ii)!! Because in (ii) it is explicity says that all the indexex $m_k,n_k>K_0$ HAVE to be in the interval $(L-\epsilon _0; L+\epsilon _0)$ !! More over let remind that (ii) is what it is given and that our assumption was concerning the existence of such $\epsilon _0$ (that is too say that $\lim_{n \to \infty}a_n\neq L$) . And we just prooved that this assumption cannot be verify in the same time that (ii) is verify. Q.E.D.
 A: You have the good feeling but a more rigorous style for $(\Leftarrow)$ would be the following:
By contradiction, let us assume that: $\lim_{n \to \infty}a_n\neq L$. It means by definition:$$\exists \epsilon>0\quad\forall N \in \mathbb{N}\quad\exists n\in \mathbb{N}\quad( n>N\text{ and }|a_n-L| \geq \epsilon)\qquad(i).$$
Let us fix such an $\epsilon.$ By assumption we know that:
$$\exists K \in \mathbb{N}\quad\forall k> K\quad|a_{n_k}-L|<\epsilon$$ and
$$\exists K'\in \mathbb{N}\quad\forall k> K'\quad|a_{m_k}-L|<\epsilon.$$
Let us now also fix such $K,K',$ and write $N_0=\max(n_K,m_{K'}).$ Since the sequence $(n_k)_k$ is increasing, we get (for all $k\in\Bbb N$) $$n_k>N_0\Rightarrow n_k>n_K\Rightarrow k>K.$$
Similarly, $m_k>N_0\Rightarrow k>K'.$ Therefore,
$$\forall k\in\Bbb N\quad(n_k>N_0\Rightarrow|a_{n_k}-L|<\epsilon)\text{ and }(m_k>N_0\Rightarrow|a_{m_k}-L|<\epsilon)\quad(ii).$$
Now according to $(i),$ there exists at least one $n>N_0$ s.t. $|a_n-L| \geq \epsilon.$ By definition, $n\in \mathbb{N}=\{ m_k\mid k\in\Bbb N\}\cup \{n_k\mid k\in\Bbb N\}.$ Such an $n$ is equal to $m_k$ or $n_k$ for some $k.$ By $(ii),$ since $|a_n-L| \geq \epsilon,$ this implies $n\le N_0.$ But this contradicts our choice of $n>N_0.$
