Is absolute convergence a topological concept?

Question

An infinite series $\Sigma_n a_n$ is said to absolutely converge if $\Sigma_n |a_n|$ converges. Absolute convergence implies convergence.

My question is, is absolute convergence a topological concept? That is to say, does there exist a topology on $\mathbb{R}$ such that $\Sigma_n a_n$ converges in that topology if and only if $\Sigma_n |a_n|$ converges in the standard topology?

Answer 1

Let $\tau$ be a topology on $\mathbb{R}$ under which every absolutely convergent series converges.

Lemma. If $(x_n) \to L$ under the standard topology, then $(x_n) \to L$ under $\tau$.

Proof. Let $(y_n)$ be any subsequence of $(x_n)$. Then there exist $(n_k)$ such that $|y_{n_k} - L| < 2^{-k}$. Now consider the sequence $(a_k)_{k=0}^{\infty}$ defined by

$$ a_k = \begin{cases} L, & k = 0 \\ y_{n_j} - L, & k = 2j-1 \text{ for some } j \geq 1 \\ L - y_{n_j}, & k = 2j \text{ for some } j \geq 1 \end{cases} $$

Then $\sum_{k=0}^{\infty} |a_k| < \infty$, and so its partial sum $s_k = \sum_{j=0}^{k} a_j$ converges under $\tau$. Since $s_{2k} = L$, it follows that $s_k \to L$ under $\tau$. Then, since $s_{2k-1} = y_{n_k}$, we have $y_{n_k} \to L$ under $\tau$.

So far, we have proved that every subsequence of $(x_n)$ has a further subsequence that converges to $L$ under $\tau$. This suffices to guarantee that $(x_n)$ converges to $L$ under $\tau$. ////

By this lemma, any series which converges conditionally under the standard topology also converges under $\tau$.

Answer 2

No.

Choose $c_n\uparrow 0$ and $d_n\downarrow 0$ such that $d_n-c_n=\frac1n$. A series $\sum a_n$ whose partial sums contain both $c_n,d_n$ for all large enough $n$ is obviously not going to converge absolutely, so your topology need to detect this by having a neighbourhood of $0$ that omits infinitely many points from $(c_n)$ or infinitely many from $(d_n)$. Say it is from $c_n$ for definiteness. But then we can construct a series $\sum b_n$ whose partial sums are precisely those omitted points from $(c_n)$. Since $c_n\uparrow 0$, $b_n=c_{s(n)}-c_{s(n-1)}$ are all positive (except the first term which is $c_{s(1)}<0$), so $\sum b_n$ is absolutely convergent, contradiction.