Uncountable Subset of Reals Generates Reals by Finite Integral Linear Combinations A question that I thought of earlier today that I couldn't quite get anywhere with.
Given an uncountable subset of the reals, $S$, is it always possible for any $r \in \mathbb{R}$ that we can take a finite subset of $S$, $\{s_1, s_2, \ldots, s_n\}$ and $\{a_1, a_2, \ldots, a_n\} \subset \mathbb{Z}$ such that $\sum_{i = 1}^n a_is_i = r?$
It seems that if $S$ includes an interval, then this problem should be pretty simple. A natural next attempt is to consider the set of irrational numbers on $(0, 1)$ but I couldn't get anywhere with that.
One of my classmates suggested viewing $\mathbb{R}$ as a vector space and treating $S$ as some sort of basis.
Any ideas or solutions?
 A: Not really. So, it's not possible to prove the statement, but it's hard to come up with an explicit counterexample. 
However, using the axiom of choice it is still possible to get one. Consider $(s_i)_{i \in I}$ a basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Since $\mathbb{R}$ is uncountable, while $\mathbb{Q}$ is countable, the set itself $I$ will be uncountable. Now take an $i_0$ in $I$ and consider 
$S= (s_i)_{ i \in I \backslash\{i_0\}}$ and $r = s_{i_0}$. You cannot write $s_{i_0}$ as a rational combination of the other $s_i$'s. 
The rational independence of real numbers can be  tricky. For instance, it is not easy to prove that the rational powers of $\pi$ are rationally independent. 
I wonder  whether sets of numbers like $2^{c^2}$ with $c$ in some Cantor type set (consisting of irrationals only )  are linearly independent over $\mathbb{Q}$. 
A: If you believe that $\Bbb R$ as vector space over $\Bbb Q$ has a basis (and it follows from the Axiom of Choice that any vector space over any field has a basis), then this is certainly false. Such a basis must clearly be uncountable (since $\Bbb Q$-vector spaces of countable dimension are countable as a set, which $\Bbb R$ is not), and being $\Bbb Q$-linearly independent, any one basis vector of such a basis is not a $\Bbb Q$-linear combination of the others (which still form an uncountable set of real numbers). Of course being a $\Bbb Z$-linear combination implies being a $\Bbb Q$-linear combination.
This argument is not constructive (since no such basis can be explicitely given), but it only depends on having an uncountable set of $\Bbb Q$-linearly independent real numbers. It might be possible to construct explicitely such an uncountable set, and get a proof not depending on the Axiom of Choice. I cannot think of such a construction right now, but it does seem likely that one could.
