Let be $\varphi:\mathbb{N}\to J$ a bijection. We say that $\sum\limits_{k\in J} a_k $ is convergent if $\sum\limits_{k=1}^{\infty}|a_{\varphi(k)}| $ converges. Our professor wrote the following statement without any further explanation or proof.
Let's assume that $\sum\limits_{k\in J}a_k $ converges, then
$\{\sum\limits_{k=1}^{n}|a_{\varphi(k)}|\mid n\in\mathbb{N}\} $ is bounded $\iff \{\sum\limits_{k\in E}|a_k|\mid E\subseteq J, ~ E\text{ is finite}\}$ is bounded.
It is intuitive that the equivalence must be true. However, our professor is ridiculously rigorous and deducts points wherever possible. So in my opinion just stating the equivalence and not giving any explanation seems handwaving if not wrong. What is the precise argument that makes the equivalence true?
Is the following argument sufficient? Maybe there are shorter proofs?
Let $E\subseteq J$ be a finite subset which contains $n$ elements. Then we define three bijections $\varphi_1:\mathbb{N} \to J\setminus E~$, $\varphi_2:[1,\cdots, m] \to E~$ and finally $\varphi:\mathbb{N}\to J~$ by $$ \varphi(k) =\begin{cases} \varphi_1(k), & 1\leq k\leq m \\ \varphi_2(k-m) & m<k . \end{cases} $$ Now let be $C\geq0$ be an upper bound of $\{\sum\limits_{k=1}^{n}|a_{\varphi(k)}|\mid n\in\mathbb{N}\}$. Then it follows $\sum\limits_{k\in E}|a_k|=\sum\limits_{k=1}^{m}|a_{\varphi_1(k)}|=\sum\limits_{k=1}^{m}|a_{\varphi(k)}|\leq C$. Hence, $\{\sum\limits_{k\in E}|a_k|\mid E\subseteq J, ~ E\text{ is finite}\}$ is bounded.
The other direction of the equivalence is indeed obvious as all intervalls of the type $[1,\cdots,n]$ are finite sets.
$\implies$ Since the set $\{\sum\limits_{i=1}^{n}|a_{\varphi(i)}|\mid n\in\mathbb{N}\}$ is bounded above, the series of real numbers $\sum_{i\geq 1}|a_{\varphi(i)}|$ is convergent. Let $E$ be a finite subset of $J$ and consider $I=\{i\geq 1: \varphi(i)\in E \}$. Since $\varphi$ is a bijection, $I$ has a greatest element, say $N$. Then $E\subset \{ \varphi(1),\ldots,\varphi(N)\}$, hence $\sum\limits_{k\in E}|a_k|\leq \sum\limits_{i=1}^{N}|a_{\varphi(i)}|\leq \sum\limits_{i=1}^{\infty}|a_{\varphi(i)}|$. Consequently, the set $\{\sum\limits_{k\in E}|a_k|\mid E\subseteq J, ~ E\text{ is finite}\}$ is bounded by $\sum\limits_{i=1}^{\infty}|a_{\varphi(i)}|$.
$\Longleftarrow$ Since $\sum\limits_{k\in J}a_k$ is convergent, by definition the series $\sum_{i\geq 1}|a_{\varphi(i)}|$ is convergent, so the set $\{\sum\limits_{i=1}^{n}|a_{\varphi(i)}|\mid n\in\mathbb{N}\}$ is bounded by $\sum\limits_{i=1}^{\infty}|a_{\varphi(i)}|$.