Can an uncountable group be generated from a single element? First question : can an uncountable group be cyclic?
Ok so my though is if $G$ is generated by i then for $x\in G$ we have $x=i^n$ for integer n, so then it must be countable. Is there a way to generate an uncountable group in some way from a single element? 
Second question: can an uncountable group be generated by a countable set?
Something related to my question is that for any positive $k: \mathbb Q^k$ is smaller than $\mathbb R$, however since we can approach any number infinitely using a binary system it makes sense to me that if we build $(N,b_1,b_2,b_3....)$ where $ N$ is the whole part of the number and b is 0 or 1 then this should have the same cardinality as $\mathbb R$.
My question is the following: can an uncountable group be generated from a countable group?

I believe that the set $2^{-n}$ for n a non negative integer does the trick in the group $\mathbb R $under addition. Am I right?

Regards
 A: If a group is generated by one element, then it is cyclic. However, all cyclic groups are isomorphic to either $\Bbb Z/n\Bbb Z$ for some $n$ or $\Bbb Z$. $\Bbb Z$ is countable, so any (infinite) group generated by one element must also be countable. We can also see this directly by simply noting that by definition, $G = \langle r\rangle = \{r^n\mid n\in\Bbb Z\}$, so there is an immediate bijection between $G$ and $\Bbb Z$.
As for the second paragraph, I'm not entirely sure what you mean. If you mean for each $b$ to be the same, then there's a bijection between $\{(N,b,b,b,\ldots)\mid N\in\Bbb Z, b\in\{0,1\}\}$ and $\Bbb Z\times\{0,1\}$ (which is countable). If you mean for each $b$ to be possibly different, just notice that if you disregard the $N$, you can put $\{(b_1,b_2,\ldots)\mid b_i\in\{0,1\}\}$ in bijection with $[0,1]$ (which is uncountable) by sending $(b_1,b_2,\ldots)\mapsto .a_1a_2\ldots$, where $.b_1b_2\ldots$ is the base $2$ decimal expansion of $.a_1a_2\ldots$, so you obtain a bijection between $\{(N,b_1,b_2,\ldots)\mid b_i\in\{0,1\}, N\in\Bbb Z\}$ and $\Bbb Z\times [0,1]$ (still uncountable).
The set $\{2^{-n}\mid n\in\Bbb Z\}$ will not generate all of $\Bbb R$ as a normal group - it won't even generate all of $\Bbb Q$ (you'll always have powers of $2$ in the denominator)! In fact, it is known in general that if a group is countably generated, it is countable. However, if you want to talk about topological groups, then you can get uncountable groups from countable generating sets, because you take the topological closure in the end. $\Bbb R$ does have a topology, so if you consider the generating set $\{10^n\mid n\in\Bbb Z\}$ as generating a topological group with the usual topology (induced by the Euclidean metric), you can indeed generate all of $\Bbb R$ (think about limits and infinite decimal expansions). However, the key here is that in $\Bbb R$ you can make sense of things like infinite sums and limits. See here for some more commentary.
A: For a countable set to generate an uncountable "algebraic" system you either need an uncountable number of operations or operations with an infinite number of operands or infinite sets as operands. (I put "algebraic" in quotes because many people would say these possibilities take you outside algebra.)
The additive subgroup of $\mathbb{R}$ generated by numbers of the form $2^{-n}$ is countable (it is a subset of the rational numbers). Yes, there are senses in which $\mathbb{R}$ is generated by these numbers, but this needs an infinitary operation such as the operation that assigns the limit to a Cauchy sequence or the operation that assigns the supremum to a non-empty bounded set.  
A: Theorem: An countably generated group is countable.
(Thus, an uncountable group cannot be countably generated.)
Proof: The proof is quite easy, in a way. Every countably generated group is the homomorphic image of the (unique) free group on countable many generators $F_{\mathbb{N}}$. So, if you can prove that $F_{\mathbb{N}}$ is countable then you are done. However, this holds as, for example, the commutator subgroup* of $F_2$, the free group on two-generators, is infinitely (and co countably) generated, and subgroups of free groups are free. Done!
*In fact, any proper normal subgroup of infinite index is infinitely generated.
Note: The proof of finitely-generated is low-level and a good exercise.
