What are the generators of $(\mathbb{R},+)$? The real numbers as a group under addition are an infinitely-generated group. I'm not sure what those generators are though (or if it even makes sense to ask the question).
For example, could we say that the generators are $1, 0.1,0.01,\dots$ as a real number can be defined as a limit $\sum_i^n w_i 10^{-i}$ for some $w$?
 A: There is no finite set, nor minimal set of generators for $(\mathbb R, +)$, hence, it is an infinitely generated (uncountable) group.
A: $\Bbb R$ is uncountable, so it definitely isn't finitely generated, or even countably generated. Suppose we had countably many generators, $\{a_{i}\}$. Then every element in $\Bbb R$ can be written as $\sum_i n_{i} a_{i}$ (let's not worry about convergence), where $i$ ranges over $\Bbb N$, and $n_{i} \in \Bbb Z$. However the set of all such expressions is clearly a countable union of countable sets, and is thus countable.
A: This hint is a good point but need you to know Divisible groups. It is not finitely generated because any nontrivial divisible group cannot be finitely generated.
A: When you talk about $\mathbb{R}$ as an abstract group, you've forgotten its topology, so you're no longer allowed to say the word "limit." The subgroup generated by the numbers $\frac{1}{10^n}, n \in \mathbb{N}$ is the subgroup of rational numbers with terminating decimal expansions, which is much smaller than the real numbers.
However, if you remember the topology, then $\mathbb{R}$ becomes a topological group, and then it's true that $\mathbb{R}$ is topologically generated by the numbers $\frac{1}{10^n}$ in the sense that the closure of the subgroup they generate is all of $\mathbb{R}$. 
In practice, though, once we start talking about topological groups it no longer becomes particularly useful to work with generators. Instead, in the special case of Lie groups a reasonable substitute is to work with their Lie algebras, which you can think of as their "infinitesimal generators." 
