# Basis for $\mathbb{R}$ over $\mathbb{Q}$

Give me some examples of basis for $\mathbb{R}$ (as vector space over field $k=\mathbb{Q}$).

Thanks.

-
The existence of such Hamel basis is ensured only by the AC(Axiom of Choice), and there seems no way to construct a concrete example of such a basis without reference to AC. I do not know much about this topic, but I believe that this is equivalent to some weaker formulation of AC. That is, the existence of a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ is independent to ZF. – Sangchul Lee Feb 24 '12 at 13:45
Please give me orders; I really enjoy being told what to do. Oh, wait. I don't. Never mind... – Arturo Magidin Feb 24 '12 at 16:54

In fact the assertion that every vector space has a basis is equivalent to the axiom of choice. This does not mean that every infinitely dimensional space has no basis. For example $\mathbb R[x]$ as a vector space over $\mathbb R$ is infinitely dimensional, but it has a basis - $\{x^n\mid n\in\mathbb N\}$.
There are models of set theory without the axiom of choice in which there is no basis for $\mathbb R$ over $\mathbb Q$, which means that one cannot just "write down" such basis, but rather that one can prove the existence of a basis in a non-constructive manner such as Zorn's lemma.
A few other such intangibles include: an explicit well-ordering on $\mathbb{R}$; a finitely additive probability measure that is not countably additive; an explicit element of $(\ell_\infty)^* \setminus \ell_1$. – Willie Wong Feb 24 '12 at 14:23
BTW, I think your second paragraph can use some clarifying. I think you mean to say: "This does not mean that a basis can never be explicitly written down for an infinite dimensional vector space." And perhaps the example of $\mathbb{R}[x]$ can be given too. – Willie Wong Feb 24 '12 at 14:35