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

Please let me know if I am on the right track or if this solution is sufficient.

My Solution:
Let $(a_n)$ be a convergent sequence. Let $\mathcal{E}>0$ be arbitrary.
Since $(a_n)$ is a convergent sequence, then $a_n \rightarrow L$, where $L$ is the limit of $(a_n)$.
This implies that there exists an $N : |a_n -L|<\mathcal{E}$ whenever $n\geq N$

Let $(a_{n_j})$ be a subsequence of $(a_n)$.
Well, $n_j \geq j \geq N$ which implies $n_j \geq N$
So the same $N$ for $(a_n)$ works for the subsequence $(a_{n_j})$ as well.
So there exists an $N : |a_{n_j}-L|<\mathcal{E}$ which implies $(a_{n_j}) \rightarrow L$

  • $\begingroup$ Looks almost fine except $n_j\geq n$ $\endgroup$ – Juniven May 12 '17 at 2:21
  • 2
    $\begingroup$ I believe I only got a small part of your comment. I must also say that $n_j \geq j \geq N$? $\endgroup$ – James Snell May 12 '17 at 3:12
  • $\begingroup$ That's correct. $\endgroup$ – Juniven May 12 '17 at 3:25

Given $\epsilon>o$, choose $N\in\Bbb N$ such that if $n\geq N$ then $|a_n-L|<\epsilon$. Therefore, if $j\geq N$ then $n_j\geq j\geq N$ which implies that $n_j\geq N$ and so $|a_{n_j}-L|<\epsilon$. Hence, $$\lim_{j\to\infty}a_{n_j}=L.$$

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