Definition:
Let be $M$ a countable set. Then $\sum\limits_{k\in M}a_k$ converges if there exists a bijection $\varphi:\mathbb{N}\to M$ such that $\sum\limits_{k=1}^{\infty}a_{\varphi(k)}$ converges absolutely. (We know that the particular choice of $\varphi$ doesn't matter)
Let's assume $M_1,M_2$ are both disjoint subsets of $\mathbb{N}$. Further, we assume that the three series $\sum\limits_{k\in M_1\cup M_2}a_k$, $\sum\limits_{k\in M_1}a_k$ and $\sum\limits_{k\in M_2}a_k$ converge. Show that
$$
\sum\limits_{k\in M_1\cup M_2}a_k=\sum\limits_{k\in M_1}a_k+\sum\limits_{k\in M_2}a_k.
$$
My new approach:
Let's define:
$a_k^{M_1\cup M_2}:=\begin{cases}
a_k & \text{ if } k \in M_1\cup M_2\\
0 & \text{ if } k\notin M_1\cup M_2,\end{cases}$
and for $i=\{1,2\}$:
$a_k^{M_i}:=\begin{cases}
a_k & \text{ if } k \in M_i\\
0 & \text{ if } k\notin M_i.\end{cases}$
There exists an index $n_0\in\mathbb{N}$ such that $\left|\sum\limits_{k=1}^{n_0}a_{\varphi(k)}-\sum\limits_{k\in M_i}a_k\right|<\frac{\epsilon}{2}$. Further, we choose $m_0\in\mathbb{N}$ large enough such that $\{\varphi(1),\varphi(2),\cdots,\varphi(n_0)\}\subseteq \{1,2,\cdots, m_0\}$ and $\sum\limits_{k=m_0}^{\infty}|a_k^{M_i}|<\frac{\epsilon}{2}$ (we know that it converges absolutely due to comparison with $\sum\limits_{k\in \mathbb{N}}a_k$). It follows:
$$\left|\sum\limits_{k=1}^{m_0}a_k^{M_i}-\sum\limits_{k\in M_i}a_k\right|\leq\left|\sum\limits_{k=1}^{m_0}a_k^{M_i}-\sum\limits_{k=1}^{n_0}a_{\varphi(k)}\right|+\left|\sum\limits_{k=1}^{n_0}a_{\varphi(k)}-\sum\limits_{k\in M_i}a_k\right|< \sum\limits_{k=m_0}^{\infty}|a_k^{M_i}|+\frac{\epsilon}{2} < \epsilon.$$
For an arbitrary $k\in\mathbb{N}$ we have $a_k^{M_1\cup M_2}=a_k^{M_1}+a_k^{M_2}$ so it follows $\sum\limits_{k=1}^{\infty}a_k^{M_1\cup M_2}=\sum\limits_{k=1}^{\infty}a_k^{M_1}+\sum\limits_{k=1}^{\infty}a_k^{M_2}\implies \sum\limits_{k\in M_1\cup M_2}a_k=\sum\limits_{k\in M_1}a_k+\sum\limits_{k\in M_2}a_k$.
Is this correct? It looks a little bit too simple… This whole thing with countable index sets still seems strange to me. Maybe there is a different approach?
Best Answer
For $i=1,2$ let $\varphi_i:\Bbb N\to M_i$ be a bijection, and let
$$\varphi:\Bbb N:M_1\cup M_2:n\mapsto\begin{cases} \varphi_1\left(\frac{n}2\right),&\text{if }n\text{ is even}\\ \varphi_2\left(\frac{n-1}2\right),&\text{if }n\text{ is odd.} \end{cases}$$
(Note that for me $\Bbb N$ includes $0$.) It’s not hard to verify that $\varphi$ is a bijection such that
$$\langle\varphi(0),\varphi(1),\varphi(2),\varphi(3),\ldots\rangle=\langle\varphi_1(0),\varphi_2(0),\varphi_1(1),\varphi_2(1),\ldots\rangle\,,$$
so that
$$\sum_{k=0}^{2n+1}a_{\varphi(k)}=\sum_{k=0}^na_{\varphi_1(k)}+\sum_{k=0}^na_{\varphi_2(k)}$$
for each $n\in\Bbb N$.
Let
$$\ell=\sum_{k\in M_1\cup M_2}a_k=\sum_{k\in\Bbb N}a_{\varphi(k)}$$
and
$$\ell_i=\sum_{k\in M_i}a_k=\sum_{k\in\Bbb N}a_{\varphi_i(k)}$$
for $i=1,2$; we want to show that $\ell=\ell_1+\ell_2$.
Let $\epsilon>0$; there is an $n_\epsilon\in\Bbb N$ such that
$$\left|\ell_i-\sum_{k=0}^{n_\epsilon}a_{\varphi_i(k)}\right|<\frac{\epsilon}2$$
for $i=1,2$. Then
$$\begin{align*} \left|(\ell_1+\ell_2)-\sum_{k=0}^{2n+1}a_{\varphi(k)}\right|&=\left|(\ell_1+\ell_2)-\left(\sum_{k=0}^na_{\varphi_1(k)}+\sum_{k=0}^na_{\varphi_2(k)}\right)\right|\\ &\le\left|\ell_1-\sum_{k=0}^na_{\varphi_1(k)}\right|+\left|\ell_2-\sum_{k=0}^na_{varphi_2(k)}\right|\\ &<\frac{\epsilon}2+\frac{\epsilon}2\\ &=\epsilon\,, \end{align*}$$
so
$$\ell=\lim_{n\to\infty}\sum_{k=0}^{2n+1}a_{\varphi(k)}=\ell_1+\ell_2\,.$$