Countable/uncountable basis of vector space

Definition: $A$ is a countable set if there exists a function $f\colon A\to\mathbb N$ which is a injective. If $A$ is not countable we say that $A$ is uncountable.

Fact I: The set $\mathbb Q$ of the rational numbers is countable.

Fact II: The set of the real numbers, $\mathbb R$ is uncountable. Furthermore, there is a bijection between the set $\mathbb R$ and $\{A\mid A\subseteq\mathbb N\}$ and the collection of sequences of natural numbers.

Fact III: If $A$ is countable then the set of all finite subsets of $A$ is countable too.

Suppose that $\mathbb R$ was of a countable dimension over $\mathbb Q$, this means that there was a countable set $\{x_i\mid i\in\mathbb N\}$ such that every real number is a linear combination over $\mathbb Q$ of a finite set of $x_i$'s.

However the collection of finite subsets of a countable set is countable. In particular we have that the span of countably many real number over $\mathbb Q$ is countable. The real numbers, however, are uncountable. Therefore the dimension cannot be countable!

I am posting this as an answer because it is too long for a comment. I thought that it might be interesting for you to know that the basis is not only uncountable, it has cardinality $\mathfrak c=2^{\aleph_0}$.

If you search for "hamel basis" "R over Q" at google or google books you'll probably find many proofs that the basis is uncountable and, moreover, that the cardinality is $\mathfrak c$.

For example:

• Proof from ask an algebraist by Henno Brandsma, link
• It is mentioned in Cieselski's Set theory for the working mathematician, p.111.

A detailed proof that the cardinality of each Hamel basis of $\mathbb R$ over $\mathbb Q$ is $\mathfrak c$ is given in Theorem 4.2.3 in Kuczma's book An introduction to the theory of functional equations and inequalities.