The multiplicative group $\Bbb Q^+$ can be viewed as a $\Bbb Z$-module. To see this, note that any rational can be decomposed into the form


$2^{n\_2} \cdot 3^{n\_3} \cdot 5^{n\_5} \cdot ...$


The tuple of coefficients $(n\_2, n\_3, n\_5, ...)$ is then an element in the module $\Bbb Z^{(\omega)}$, the set of integer sequences with finite support.


The dual module $Hom(\Bbb Q^+, \Bbb Z)$ mapping from rationals to integers is $\Bbb Z^\Bbb N$, the set of *all* integer sequences. For various reasons in musical tuning theory, an important class of these sequences are given by


$f\_r = (\lfloor r \cdot \log(2) \rceil, \lfloor r \cdot \log(3) \rceil, \lfloor r \cdot \log(5) \rceil, ...)$


for $r \in \Bbb R$, and where $\lfloor x \rceil$ rounds x to the nearest integer.


It has sometimes proven useful to put the Euclidean topology on these functionals, because then we can find local maxima with respect to various functions on them.


This brings me to my questions:


1. Is it possible, in a natural way, to extend the topology on the set of $f\_r$ to the set of all finite $\Bbb Z$-linear combinations of that set?
2. Likewise, is it possible to extend the topology for infinite $\Bbb Z$-linear combinations, when such sums converge?
3. We can also treat the above as a discrete lattice of vectors embedded in a real vector space. Is it possible to extend the topology on the $f\_r$ to the $\Bbb R$-linear span of finite weighted combinations?
4. Likewise, is it possible to extend the topology to infinite $\Bbb R$-linear combinations, when they converge?


I have a hunch one may be able to use the product topology for this, but am not quite sure.



Extending the topology on a set to the group/vector space it generates
The answer to your questions 1 and 3 is that there is a universal way to do it, but I don't know how this works for infinite linear combinations.


Unfortunately, I only know how to do it in the category of compactly generated topological spaces (or any cartesian closed category of spaces, really). Hence by a topological ring I mean a ring $R$ equipped with a compactly generated topology making the operations continuous with respect to the compactly generated product (and similarly for topological $R$-modules).


My claim is that the forgetful functor from the category of topological $R$-modules to the category of (compactly generated) spaces has a left adjoint. Here is an explicit construction. Let $X$ be a compactly generated space and let $R X$ be the set of all $R$-linear combinations of the points of $X$. We topologize it as follows. For each $m \in \mathbb{N}$ we consider the function $\phi\_m \colon R^m \times X^m \to R X$ defined as $\phi\_m(a, x) = a\_1 x\_1 + \ldots + a\_m x\_m$. We take the finest topology on $R X$ making $\phi\_m$ continuous for all $m$.


It is quite routine to verify that $R X$ satisfies the universal property of a free topological $R$-module on $X$. The only wrinkle appears when we check the continuity of the addition $R X \times R X \to R X$. Here we need to use the fact that products preserve colimits in order to show that the topology of $R X \times R X$ is generated by the maps $\phi\_m \times \phi\_n$.


If we work with all topological spaces, then the forgetful functor still has a left adjoint as follows by the Adjoint Functor Theorem. However, in this case I don't know how to explicitly describe the resulting topology. The construction above will sometimes fail, e.g. if $X$ is discrete of cardinality $\mathfrak{c}$ then the addition map on $\mathbb{R} X$ is discontinuous with respect to the standard product topology.


Note: this answer is based on a discussion from [nForum](https://nforum.ncatlab.org/discussion/6827/monadicity-of-topological-vector-spaces/).


