Subsequences of a convergent sequence converge to the same limit as the original sequence

Question

Here is my proof:

Proof. Let $(a_j) \to L$ be the original sequence and $({a}_{n_{j}})$ be a subsequence of $(a_j)$. We have to then show that $ ({a}_{n_{j}}) \to L$.

Since $(a_j) \to L$, we have for all $\epsilon >0$, there exists $N \in \mathbb{N}$ such that $|a_j - L| <\epsilon $ for all $j\ge N$. But $n_j \ge j$ which implies $n_j \ge N$ and it follows that $|a_{n_j}-L|<\epsilon$. Hence, $ ({a}_{n_{j}}) \to L$. $\Box$

Is my proof correct? I'm not sure why it is correct to claim $n_j \ge j$ but it seems intuitive.

Answer 1

Your proof is correct. In particular, $n_j \geq j $ is true because $n: \mathbb {N} \to \mathbb {N} $ is a strictly increasing function by definition of subsequence. You can give an easy proof using induction.

Answer 2

Since a subsequence must be infinite subset of the original sequence whose elements are ordered in same way as the sequence. So, the indices $(n_j)$ must satisfy $n_j > n_i$ if $j > i$ and also $n_j \rightarrow \infty$. For your question you can check it by using induction,

Clearly $n_1 \ge 1$. Suppose $n_{j-1} \ge j-1$, then $n_j > n_{j-1} \ge j-1$, that is $n_j \ge j$.