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Let $\sigma$ be a bijection of $\mathbb{N}$ onto itself, and for each n, let $\sigma(n)$ be the smallest number of intervals $[a, b]$ in $\mathbb{N}$ such that the union of these intervals is $\sigma([0, n])$.

(a) Suppose $\sigma$ is bounded in $\mathbb{N}$. Let $(x_n)$ be a convergent series in a normed space $E$, show that the series $(x_{\sigma(n)})$ is convergent in $E$ and that $$\sum_{n=0}^{\infty} x_{n}=\sum_{n=0}^{\infty} x_{\sigma(n)}$$

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  • $\begingroup$ $\sigma$ cannot be both bounded and a bijection of $\mathbb{N}$ onto itself. Also, I don't understand what your definition of $\sigma(n)$ means. It is, at very least, self-referential. $\endgroup$ – nullUser Jun 4 '13 at 17:52
  • $\begingroup$ I also didn't understand the definition of $\sigma(n)$. However I think this is saying $\sigma$ is a bounded operator in the sense that $\sigma(n)\leq Bn$ for some $B<\infty$; this is not contradictory with being a bijection, as opposed to being bounded in the sense that $\sigma(n)<B$ for all $n$. $\endgroup$ – Laura Balzano Jun 4 '13 at 19:25
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I'll give an answer to this: Let $\sigma$ be a bijection of $\mathbb{N}$ onto itself which is bounded in $\mathbb{N}$. Let $x_n$ be a convergent series in a normed space $E$, show that the series $x_{\sigma(n)}$ is convergent in $E$ and their limits are equal. This answer may be sketchy in some steps so please confirm things for yourself.

Define $s = \sum_{n=0}^\infty x_n.$ Let $\epsilon>0$. We should find $N$ such that for all $m\geq N$, $$\left\|\sum_{n=0}^m x_{\sigma(n)} - s \right\| < \epsilon.$$ First we find $\tilde{N}$ such that $$\left\|\sum_{n=0}^m x_n - s \right\| < \epsilon \implies \left\|\sum_{n=m+1}^\infty x_n\right\|<\epsilon$$ for all $m \geq \tilde{N}$, which we can do because $x_n$ is convergent. Since $\sigma$ is a bounded mapping, $\sigma(n) \leq Bn$ for some $B<\infty$. Take $N = B\tilde{N}$. Therefore for all $n \leq \tilde{N}$, $\sigma(n) \leq N$. Let $\mathcal{A}_m = \{ n > \tilde{N} : \sigma(n) \leq m \}$; Note that $\{1,\dots,\tilde{N}\} \notin \mathcal{A}_m$ since $\sigma$ is 1-1. So in particular, $$\sum_{n=0}^N x_{\sigma(n)} = \sum_{n=0}^{\tilde{N}} x_n + \sum_{n \in \mathcal{A}_N} x_n$$ and for $m>N$, $$\sum_{n=0}^m x_{\sigma(n)} = \sum_{n=0}^{\tilde{N}} x_n + \sum_{n \in \mathcal{A}_m} x_n .$$ We conclude: $$\left\|\sum_{n=0}^m x_{\sigma(n)} - s \right\| = \left\| \sum_{\substack{ n=\tilde{N}+1 \\ n \notin \mathcal{A}_m}}^\infty x_n \right\| <\epsilon.$$

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