Cardinality of the basis for $\mathbb{R}$ over $\mathbb{Q}$? This question came up as a discussion I had with a friend. Clearly, the basis is not of finite cardinality since that would imply the set $\mathbb{R}$ has the cardinality $\aleph_{0}$ which is false. Therefore the basis clearly has to be infinite. However, the question that came up was that is the basis countably infinite or uncountably infinite. So I'd like to know the cardinality of the basis and also appreciate some pointers to the proof. 
Thanks in advance. :)
 A: The size of the basis cannot exceed $\aleph$, as the set $\mathbb{R}$ spans your vector space.
The size of the basis is indeed $\aleph$. To see this, suppose we use a basis of size $\aleph_0$ (or any other cardinality less then $\aleph$ , for that matter). Let us denote this basis by ${x_1,x_2,x_3,...,x_n,...}$. Any element of $\mathbb{R}$ can be written as a finite linear combination of the form $\sum_{i=1}^{M} q_{i}x_{r_i}$ where $q_{i}\in\mathbb{Q}$.
In particular, if we fix some element $y\in\mathbb{R}$, all of the indices ${r_i}$ appearing in the linear combination are no more then some $N$, meaning that we can write $y$ as a linear combination of $x_1 ,..., x_N$. 
Because $y\in\mathbb{R}$ was arbitrary, we conclude that:
$$
\mathbb{R}\subseteq \bigcup_{N=1}^{\infty} span_\mathbb{Q} (x_1,...,x_N)
$$
As you said yourself, each of the sets $span_\mathbb{Q} (x_1,...,x_N)$ is countable, meaning that the union $ \bigcup_{N=1}^{\infty} span_\mathbb{Q} (x_1,...,x_N)$ has cardinality $\aleph_0$, so it cannot contain $\mathbb{R}$.
A: For infinte dimensions, the cardinality of a vector space over $\mathbb Q$ equals the cardinality of the base.  
A: Since any element fo $\;\Bbb R\;$ has to be a finite $\;\Bbb Q$- linear combination of some basis, the basis can not be countable, either. Thus...
