Let $x_n$ be a sequence such that $x_n \rightarrow 0$. Let $\sigma\colon\mathbb N \rightarrow \mathbb N$ be a bijection. Define a new sequence $y_n:= x_ {\sigma (n)} $. Show that $ y_n \rightarrow 0 $.

ATTEMPT Since $x_n$ converges to 0 implies after a certain $n>n_{0}$ all its terms will lie between $(0-\epsilon,0+\epsilon$). As $\sigma$ is a bijection so it is increasing function. If I take set of subscripts of original sequence, I take all $n$ after $n>n_0$ such that set $\{ n_1 ,n_2,n_3,n_4,\ldots\}$ where all $n_{i} ,i=1,2,3,\ldots$ are bigger than $n_0$. So applying $\sigma$ function which is an increasing function I get new increasing sequence of subscripts. Now I need to claim that $x_i = \sigma (n_i)$ in order to make furthure progress. But how? Thanks

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    $\begingroup$ A bijection doesn't have to be necessarily increasing. $\endgroup$ – Anurag A May 12 '15 at 6:09
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    $\begingroup$ I took the liberty of giving the question a less symbol-heavy title $\endgroup$ – Trevor Wilson May 12 '15 at 6:29
  • $\begingroup$ @TrevorWilson can you explain title $\endgroup$ – Taylor Ted May 12 '15 at 6:33
  • $\begingroup$ is it true that only bijections are of form sigma(n) = kn ,where k is even $\endgroup$ – Taylor Ted May 12 '15 at 6:35
  • $\begingroup$ A permutation is a bijection from a set to itself (such as $\sigma$.) So the question asks: if we take a sequence that converges to zero, and we use a permutation to rearrange its terms, does it still converge to zero? (And there is nothing special about zero here.) $\endgroup$ – Trevor Wilson May 12 '15 at 6:35

Intuitively: The property that there exists $n_0$ such that $|x_n|<\epsilon$ for all $n>n_0$ can be reformulated as: The number of terms outside the interval $(-\epsilon,\epsilon)$ is finite. In this reformulation we see that the order of $\mathbb N$ doesn't matter.

Explicitly, one can proceed as follows: Given $\epsilon>0$, there exists $n_0$ such that $|x_n|<\epsilon$ for all $n>n_0$. Let $n_1=\max\{\sigma^{-1}(1),\ldots,\sigma^{-1}(n_0)\}$. Then for all $n>n_1$, $\sigma(n)$ is not in $\{1,\ldots,n_0\}$, hence $|y_n|=|x_{\sigma(n)}|<\epsilon$.

  • $\begingroup$ i couldnot understand about sigma inverses here $\endgroup$ – Taylor Ted May 12 '15 at 6:14
  • $\begingroup$ $\sigma$ is a bijection, so $\sigma^{-1}$ is the inverse permutation. $\endgroup$ – Josh Chen May 12 '15 at 6:38

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