Here's the statement I want to prove:
Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of real numbers that converges to a real number $L$. Then, every subsequence $\{a_{n_k}\}_{k=1}^{\infty}$ converges to $L$.
Proof Attempt:
Let $\epsilon > 0$ be arbitrary but fixed. We are required to prove that:
$$\exists K \in \mathbb{N}: \forall k \geq K: |a_{n_k}-L| < \epsilon$$
We know that there exists an $N_0 \in \mathbb{N}$ such that:
$$\forall n \geq N_0: |a_n-L| < \epsilon$$
Since $\{n_k\}_{k=1}^{\infty}$ is a strictly increasing sequence of natural numbers, then:
$$\exists K \in \mathbb{N}: \forall k \geq K: n_k \geq N_0$$
$$\implies \exists K \in \mathbb{N}: \forall k \geq K: |a_{n_k}-L| < \epsilon$$
which is exactly the assertion that $\lim_{k \to \infty} (a_{n_k}) = L$. That proves the desired result.
Is the proof above correct? If it isn't, why? How can I fix it?
Best Answer
Your proof is correct. In fact, you could use your proof to derive a method to find an explicit suitable $K$ for each $\epsilon$, for the subsequence, given a method for the sequence itself.