Is my understanding of the proof of rearrangement theorem correct? I'm reading the proof of rearrangement theorem. Could you please verify if my understanding of the last part of the proof is correct or not?






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*The authors said



The inequality $(8.3)$ implies the absolute convergence of $\sum x_{\sigma(k)}$.

I think the inequality $(8.3)$ implies that $\sum_{k=0}^{m}|x_{\sigma(k)}| \le \sum_{k=0}^{N}|x_{k}| + \varepsilon$ for all $m \ge M$. As such, the sequence $(\sum_{k=0}^{m}|x_{\sigma(k)}|)_{m \in \mathbb N}$ is bounded and thus it converges. Since $(\sum_{k=0}^{m}|x_{\sigma(k)}| )_{m \in \mathbb N}$ converges, $\sum x_{\sigma(k)} := (\sum_{k=0}^{m} x_{\sigma(k)})_{m \in \mathbb N}$ converges absolutely.


*The authors said



We see that $\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{N} x_{k}\right| \leq \varepsilon$, so the values of the two series agree.

I think because $\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{N} x_{k}\right| \leq \varepsilon$ holds, $\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{n} x_{k}\right| \leq \varepsilon$ holds for all $n \ge N$. As such, we take the limit  $n \to \infty$ and get $$\left|\sum_{k=0}^{\infty} x_{\sigma(k)}- \lim_{n \to \infty}\sum_{k=0}^{n} x_{k}\right| \leq \varepsilon \quad \text{or equivalently} \quad \left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{\infty} x_{k}\right| \leq \varepsilon$$
We again take the limit  $\varepsilon  \to 0$ and get $$\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{\infty} x_{k}\right| \le \lim_{\varepsilon  \to 0}\varepsilon = 0  \quad \text{or equivalently} \quad \sum_{k=0}^{\infty} x_{\sigma(k)} = \sum_{k=0}^{\infty} x_{k}$$
As such, the values of the two series agree.
 A: *

*Correct.

*Let $S:=\sum_{k=0}^\infty x_{\sigma(k)}$. Then $\big| S-\sum_{k=0}^n x_k\big|\le\varepsilon$ will indeed follow for $n>N$, but your reasoning is not enough. 
Use that, just as for $N$, there's also an $M'(\ge M)$ for $n$, such that all indices $0,1,\dots,n$ are present within $\sigma(0),\dots,\sigma(M')$.
Once it's proved, we can simply finish by the definition of limit, as for every $\varepsilon>0$ we have an $N$ such that $n\ge N$ implies $|S-\sum_{k=0}^n x_k|\le\varepsilon$. 
A: From @Berci's answer, I fixed my attempt here.

My attempt:
For each $\varepsilon>0$ , there is some $N \in \mathbb{N}$ such that $\sum_{k=N+1}^{m}|x_{k}|<\varepsilon$ for all $m>N$. Taking the limit $m \rightarrow \infty$ yields the inequality $\sum_{k=N+1}^{\infty}|x_{k}| \leq \varepsilon$.
Let $\sigma$ be a permutation of $\mathbb{N}$. For each $n \ge N$, let $p_n =\max \{\sigma^{-1}(0), \ldots, \sigma^{-1}(n)\}$ we have $\{\sigma(0), \ldots, \sigma(p_{n})\} \supseteq\{0, \ldots, n\}$. Then we have $$\left|\sum_{k=0}^{m} x_{\sigma(k)}-\sum_{k=0}^{n} x_{k}\right| \leq  \sum_{k=n+1}^{\infty}\left|x_{k}\right|\leq  \sum_{k=N+1}^{\infty}\left|x_{k}\right| \leq \varepsilon, \quad m \geq p_n, n \ge N \tag{8.2}$$ and also $$\left|\sum_{k=0}^{m} |x_{\sigma(k)}|-\sum_{k=0}^{n}| x_{k}| \right| \le \left|\sum_{k=n+1}^{\infty} |x_k| \right| = \sum_{k=n+1}^{\infty} |x_k| \le \sum_{k=N+1}^{\infty} |x_k| \leq \varepsilon, \quad m \geq p_n, n \ge N \tag{8.3}$$
In the particular case $n=N$, the inequality $(8.3)$ becomes $\left|\sum_{k=0}^{m} |x_{\sigma(k)}|-\sum_{k=0}^{N}| x_{k}| \right| \le \varepsilon$ for all $m \ge p_N$. This implies $\sum_{k=0}^{m}|x_{\sigma(k)}| \le \sum_{k=0}^{N}|x_{k}| + \varepsilon$ for all $m \ge p_N$. Consequently, the sequence $(\sum_{k=0}^{m}|x_{\sigma(k)}|)_{m \in \mathbb N}$ is bounded and thus converges. Since $(\sum_{k=0}^{m}|x_{\sigma(k)}| )_{m \in \mathbb N}$ converges, $\sum x_{\sigma(k)} := (\sum_{k=0}^{m} x_{\sigma(k)})_{m \in \mathbb N}$ converges absolutely.
We take the limit $m \to \infty$ in $(8.2)$ and get $$\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{n} x_{k}\right| \leq  \varepsilon, \quad n \ge N \tag{8.4}$$
We continue to take the limit $n \to \infty$ in $(8.4)$ and get $$ \quad \left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{\infty} x_{k}\right| \leq \varepsilon$$
Finally, we take the limit  $\varepsilon \to 0$ in the above inequality and get  $$\sum_{k=0}^{\infty} x_{\sigma(k)} = \sum_{k=0}^{\infty} x_{k}$$
As such, the values of the two series agree.
