Prove or disprove $ lim_{n \to \infty} a_n = lim_{n \to \infty} a_{\pi(n)} $, where $\pi$ is a bijective function

Question

I have no idea how to solve the following task:

Prove or disprove that: If $ (a_n)_{n \in \mathbb{N}} $ converges and $ \pi : \mathbb{N} \mapsto \mathbb{N} $ is a bijective function, then $$ lim_{n \to \infty} a_n = lim_{n \to \infty} a_{\pi(n)}. $$

Since this is my homework, I'm not asking for a solution. Could you just give me some hints how to solve this? I think I should use the properties of a bijective function here, but I have no idea, how... Any help would be appreciated!

Regards, Lena

Related http://math.stackexchange.com/questions/185778/rearrangement-of-sequences-with-limit-0 — clark, Jan 16 '14 at 23:25

score 1 · Accepted Answer · answered Jan 16 '14 at 23:24

1

Hint: Any open interval around the limit contains all but finitely many elements of the sequence.

answered Jan 16 '14 at 23:24

Berci

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Prove or disprove $ lim_{n \to \infty} a_n = lim_{n \to \infty} a_{\pi(n)} $, where $\pi$ is a bijective function

1 Answers1