Extending the topology on a set to the group it generates 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:


*

*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?

*Likewise, is it possible to extend the topology for infinite $\Bbb Z$-linear combinations, when such sums converge?

*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?

*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.
 A: Further everywhere I’ll suppose that if we extend a topology from a subset of a group to the whole group then the extended topology should be a group topology, that is making multiplication and inversion on the group continuous. 
It seems that the family $F=\{f_r:r>0\}$ is linearly independent over $\Bbb Q$. If so then the set $\langle F\rangle$ of all finite $\mathbb Z$-linear combinations of the set $F$ contains a family $F’=\{f_r:r\in\Bbb R\}$ and is a free abelian group $A(F)$ generated by the set $F$. The group $A(F)$ admits a canonical group topology, that is a topology of a free abelian topological group of the space $F$, which is the strongest group topology on $A(F)$, extending the topology of the space $F$ (see, for instance, [AT, Th. 7.1.2]). Thus the free abelian topological group $A(F)$ is naturally isomorphic to free abelian topological group $A(\Bbb R)$ over the reals. The topology of the group $A(\Bbb R)$ is not very good: I didn't checked references, but I guess that, conversely to Euclidean topology, is neither locally compact nor first countable, but still $k_\omega$. I suspect that there is no Euclidean (finitely-dimensional) group topology on $A(F)$ extending the topology of $F$ (or $F’$). 
To extend topology from $\langle F\rangle$ further to a larger group you can simply claim the set $\langle F\rangle$ open (as far as I know, other ways are quite complicated and there is no general method of topology extension, but only ideas working for specific cases). 
Concerning questions 3 and 4 you can also use a conception of a free linear vector $L(F)$ space over a topological space $F$ (I guess that the set $F$ is linearly independent over $\Bbb R$ too), but I am not familiar with this construction.
At last, I remark that the topology which the set $F$ inherits from the product $\Bbb Z^\Bbb N$ is, clearly, stronger than Euclidean, but does not coincide with it. To see the last remark that no Euclidean neighborhood of the point $f_{1/(2\log 2)}$ is contained in its product neighborhood $\{1\}\times\Bbb Z^{\Bbb N\setminus\{1\}} $.
References
[AT] Alexander V. Arhangel'skii, Mikhail G. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.
