# Is absolute convergence a topological concept?

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?

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 $$\tau$$.

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.

• What if we allow enlarging $\mathbb{R}$ under the new topology? Couldn't $c_n$ converge to something like $0_-$ and $d_n$ to $0_+$? – Bananach Nov 9 '18 at 6:42