# Box topology defines a topological vector space?

Is the set of all real summable sequences, endowed with the box topology, a topological vector space?

Formally, I am interested in $X=\{(a_1,\ldots):a_n \in \mathbb{R} \ \forall n, \ \sum a_n < \infty\}$ endowed with the box topology.

I think that this is the case as vector addition and scalar multiplication both seem continuous to me.

But, there is at least one place which suggests that the box topology generally does not define a topological vector space:

Why is the box topology on a product of complete topological vector spaces complete?

The problem is that multiplication by $x \in X$ is not continuous from $\mathbb R$ to $X$ if $x$ has infinitely many nonzero terms. For example, take box neighbourhood $U = \prod_{n=1}^\infty (1/n^2-1/n^4, 1/n^2+1/n^4)$ of $x$ with $x_n = 1/n^2$. There is no $\delta > 0$ such that $(1-\delta, 1+\delta) x \subseteq U$, as this would require $\delta < 1/n^2$ for all $n$.
• Well, can you? ${}$ Jan 18, 2018 at 4:01
• Hi, I see it now. Thanks very much. Is there an interesting finest topology? More specifically, I am interested in a topology where I have a topological vector space and the set of positive sequences $(a_1,...)$ s.t $\frac{a_1}{a_2}>r_1$, $\frac{a_2}{a_3}>r_2$,... is open where the $r_i$ are some constants between 0 and 1.