# Rearrangement of a Sequence

• Oct 22nd 2008, 10:22 PM
Jes
Rearrangement of a Sequence
Hi, I want to prove that a converging sequence will still converge to the same limit even if all the terms are permuted. I figured the problematic terms are the first $\displaystyle N-1$ terms since they can be located anywhere in the permuted sequence. To get rid of them, I want to find the largest permuted index of these finite terms and define that as the new starting point for the tail sequence. Here's what I got:

Suppose $\displaystyle \{a_n\}$ converges to $\displaystyle L$. Let $\displaystyle \phi : \mathbb{N} \rightarrow \mathbb{N}$ be a bijection. Define a new sequence by $\displaystyle \{ a_{\phi(n)} \}$. We want to show this new sequence converges to $\displaystyle L$.

Given any $\displaystyle \epsilon > 0$, there is some $\displaystyle N' \in \mathbb{N}$ such that if $\displaystyle n \geq N'$, then $\displaystyle d(a_n, L) < \epsilon$. Define $\displaystyle N = \max \{\phi(1), \phi(2), \ldots, \phi(N'), N' \}$ and suppose $\displaystyle \phi(n) \geq N$. Then $\displaystyle d(a_{\phi(n)}, L) \leq \epsilon$.

It seems like that should do it but I'm always skeptical when it seems easy. Thanks.
• Oct 23rd 2008, 12:07 AM
Opalg
That looks like a good proof to me.