Solution verification of Exercise 8.2.6 Terence Tao Vol1 I will use the lemma 8.2.7. I have written proof of this lemma, and want to verify, any comment or suggestion will be helpful.
Lemma 8.2.7. Let $\Sigma_{n = 0}^{\infty}a_n$ be a series of real numbers which is conditionally convegent, but not absolutely convergent. Define the sets $A_+ := \{n \in \mathbb{N} : a_n \geq 0\}$ and $A_{-} := \{n \in \mathbb{N}: a_n < 0 \}$ Thus $A_{+} \cup A_{-} = \mathbb{N}$ and $A_{+} \cap A_{-} = \emptyset$. Then both of the series $\Sigma_{n \in A_{+}} a_n$ and $\Sigma_{n \in A_{-}} a_n$ are not absolutely convergent.
Solution:
By Lemma 8.2.7 we know that $\Sigma_{n \in A_{+}} a_n$ is divergent. We are dealing with positive elements,so for $n \in A_{+}$ we have that $\Sigma_{n \in A_{+}} a_n = \infty$. We know that $A_{+}$ is infinite, so we have increasing bijection $f : \mathbb{N} \rightarrow A_{+} \subset \mathbb{N}$, therefore it follows:
$$\Sigma_{j \in \mathbb{N} : j \in A_{+}} a_j = \Sigma_{j = 0}^{\infty} a_{f(j)} = \infty $$
Thanks in advance.
 A: Proposition $8.2.6(c)$ in the book states that if $X = X_{1}\cup X_{2}$ for some disjoint sets $X_{1}$ and $X_{2}$ and $f:X\rightarrow\mathbb{R}$ then $\sum_{x\in X}f(x)$ is absolutely convergent if and only if $\sum_{x\in X_{1}}f(x)$ and $\sum_{x\in X_{2}}f(x)$ are absolutely convergent and
$$\sum_{x\in X}f(x) = \sum_{x\in X_{1}}f(x)+\sum_{x\in X_{2}}f(x).$$
Suppose both $\sum_{n\in A_{+}}a_{n}$ and $\sum_{n\in A_{-}}a_{n}$ are absolutely convergent then using the proposition stated above $\sum_{n\in\mathbb{N}}a_{n}$ is absolutely convergent and using the identity map $\text{id}:\mathbb{N}\rightarrow\mathbb{N}$ we have $\sum_{n= 
 0}^{\infty}a_{n}$ is absolutely convergent which is a contradiction (note that I have used the definition for series on countable sets which states that if $X$ is countable and $f:X\rightarrow\mathbb{R}$ and if $\sum_{x\in X}f(x)$ is absolutely convergent then $\sum_{x\in X}f(x) = \sum_{n = 0}^{\infty}f(g(n))$ for any bijection $g:\mathbb{N}\rightarrow X$).
Suppose we assume that exactly one of the series, say $\sum_{n\in A_{+}}a_{n}$ is absolutely convergent. Let
$$\sum_{n\in A_{+}}a_{n} = \sup\left\{\sum_{n\in B}a_{n}:B\subseteq A_{+},\:B\:\text{finite}\right\} = L.$$
Since $\sum_{n\in A_{-}}a_{n}$ is not absolutely convergent for the real number $M-L$ (where $M$ is some arbitrary real number) there exists a finite subset $C$ of $A_{-}$ such that
$$\sum_{n\in C}|a_{n}| = -\sum_{n\in A_{-}}a_{n} > -(M-L)\implies \sum_{n\in C}a_{n} < M-L.$$
Since $C$ is finite there exists a natural number $N$ such that $C\subseteq\{n\in\mathbb{N}:0\leq n\leq N\}$. Therefore,
$$\sum_{n = 0}^{N}a_{n} = \sum_{n\in\{n\in\mathbb{N}:0\leq n\leq N\}\cap A_{+}}a_{n}+\sum_{n\in\{n\in\mathbb{N}:0\leq n\leq N\}\cap A_{-}}a_{n}\\
\leq L+\sum_{n\in\{n\in\mathbb{N}:0\leq n\leq N\}\cap A_{-}}a_{n}<L+\sum_{n\in C}a_{n}< L+(M-L) < M.$$
Therefore, the partial sums of $\sum_{n = 0}^{\infty}a_{n}$ are not bounded below which implies that $\sum_{n = 0}^{\infty}a_{n}$ is not conditionally convergent which is a contradiction.
