Say I have a sequence $\langle a_n\rangle \subset \mathbb{R}$ with $\sum |a_n| < \infty$.
I need to show, $\forall \xi : \mathbb{N} \to \mathbb{N}$: bijection,
$$ \sum a_{\xi(n)} = \sum a_n $$
My attempt
Let $S_n = \sum |a_n|$, and since it converges, $\exists \sup \{ S_n : n \in \mathbb{R} \} = M$. Since $S_n$ : monotone increasing, $\lim_n S_n = M$.
Say $T_n = \sum |a_{\xi(n)}|$, and let $m_n = \max \{ \xi(i) : i=1,2, \cdots, n\}$, then $T_n \le S_{m_n} \le M$, thus $\exists \sup \{ T_n : n \in \mathbb{R} \} = M$.
But the most tricky part is, how to prove,
$$ \sum a_{\xi(n)} = \sum a_n $$
Is there any help?
Let $\epsilon >0$. There exists $N$ such that $\sum_{n=N+1}^{\infty} |a_n| <\epsilon /2$. If $m >N$ then $|\sum_{n=1}^{\infty} a_n-\sum_{n=1}^{m} a_n|=|\sum_{n=m+1}^{\infty} a_n|<\epsilon /2$. Also there exists an integer $k$ such that $\{1,2,...,N\} \subset \{\xi (1),\xi (2),..,\xi (k)\}$. Now suppose $m>\max \{N,k\}$. Then $$|\sum_{n=1}^{\infty} a_{\xi (n)}-\sum_{n=1}^{m} a_{\xi (n)}|=|\sum_{n=m+1}^{\infty} a_{\xi (n)}| \leq \sum_{n=N+1}^{\infty} |a_n|<\epsilon /2.$$ Finally, $$|\sum_{n=1}^{\infty} a_{\xi(n)}-\sum_{n=1}^{\infty} a_n|$$ $$\leq |\sum_{n=1}^{m} a_{\xi(n)}-\sum_{n=1}^{m} a_n|+|\sum_{n=m+1}^{\infty} a_{\xi(n)}-\sum_{n=m+1}^{\infty} a_n|.$$ You can see that that that the last sum is less than $\epsilon$. For the first sum here note that the terms $a_{\xi(n)}$, $1\leq n \leq m$ include the numbers $a_1,a_2,...,a_N$. Cancel these out and you will see that the $|\sum_{n=1}^{m} a_{\xi(n)}-\sum_{n=1}^{m} a_n| <\epsilon$.