Your proof aims to show that if a sequence $(a_n)$ converges to a limit $A$, then any subsequence $(a_{n_k})$ of $(a_n)$ also converges to the same limit $A$.

Let’s go through your proof step-by-step:

1. Main Idea: The proof leverages the definition of convergence. Specifically, if $(a_n)$ converges to $A$, then for any $\varepsilon > 0$, there exists an index $n_0$ such that for all $n > n_0$, $a_n$ is within an $\varepsilon$-neighborhood of $A$, denoted by $U_\varepsilon(A)$.

2. Structure of the Proof:

   • You start by stating that for every $\varepsilon > 0$, there exists an $n_0 \in \mathbb{N}$ such that for all $n > n_0$, $a_n \in U_\varepsilon(A)$.
   • Then, for any subsequence $(a_{n_k})$, since $n_k$ is an increasing sequence of indices, there exists some $k_0$ such that $n_k \ge n_0$ for all $k \ge k_0$.
   • Consequently, for $k \ge k_0$, $a_{n_k} \in U_\varepsilon(A)$ holds as well, meaning the terms of the subsequence eventually stay within $\varepsilon$ of $A$.
   • This implies $\lim_{k \to \infty} a_{n_k} = A$.

3. Conclusion:

   • Since we showed that for any $\varepsilon > 0$, there exists an index $k_0$ such that $a_{n_k}$ remains within $\varepsilon$ of $A$ for all $k \ge k_0$, we conclude that $\lim_{k \to \infty} a_{n_k} = A$, as required.

Summary

Your proof is correct. The argument effectively uses the fact that any subsequence inherits the limit behavior of the original sequence, as the convergence criterion holds for all sufficiently large indices in the subsequence as well.

Would you like further details, or do you have any specific questions?

Here are 5 related questions to deepen understanding:

1. Why does convergence of the original sequence guarantee the convergence of all its subsequences?
2. Can a sequence have different limits for different subsequences if it does not converge as a whole?
3. How does this proof differ when dealing with divergent sequences?
4. What happens if the sequence converges to infinity? Do subsequences also converge to infinity?
5. How would the proof change if we used a bounded sequence that does not converge?

Tip: Always remember that convergence of the whole sequence implies convergence of every subsequence to the same limit, but the converse is not necessarily true.