Let $\{a_n\}$ be a sequence and L a real number such that $\lim_{n\to\infty} a_n = L$ Prove that if $\{a_{n_k}\}$ is any subsequence of $\{a_n\}$, then $\lim_{k\to\infty} a_{n_k} = L $
I have found the following proof:
Let $\epsilon > 0$ be arbitrary, $|a_n -L| < \epsilon$ and $|a_{n_k} -L| < \epsilon$ $N>0, n \geq N$
At some $k>N$ $a_n=a_{n_k}$
Therefore, the limit of $a_{n_k}$ is equal to the limit of $a_n$
Now, is there an alternative proof for this that is simpler or more "complete" than the one I have shown?
You want to prove that $a_{n_k} \to L$, which means, that given $\epsilon >0$ there exist $k_0 \in \Bbb{N}$ such that $k \geq k_0$ implies $|a_{n_k}-L|< \epsilon$ You know that for that given $\epsilon$ there exist $N \in \Bbb{N}$ such that $n \geq N$ implies $|a_{n}-L|< \epsilon$ So in order to be sure that $|a_{n_k}-L|< \epsilon$ you have to take your $k_0$ such that $k \geq k_0$ implies $n_k \geq N$