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

Thanks.

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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. –  sos440 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

1 Answer 1

While it is perfectly reasonable to write down a basis for a finitely dimensional vector space, it is not always possible to write one for infinitely dimensional vector spaces.

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.

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I quote from the introduction of Schechter's Handbook of Analysis and its Foundations: "However, as I searched through the literature, I was unable to find explicit examples of several important pathological objects, which I now call intangibles ... I could not understand the dearth of examples until I accidentally ventured beyond the traditional confines of analysis. I was surprised to learn that the examples of these mysterious objects are omitted from the literature because they must be omitted: Although the objects exist, it can also be proved that explicit constructions do not exist." –  Willie Wong Feb 24 '12 at 14:21
    
@Willie: Well, Axiom of determinacy implies that not only an explicit construction does not exist, but that this object itself [Hamel basis] does not exist. :-) –  Asaf Karagila Feb 24 '12 at 14:23
    
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
    
@Willie: Indeed, all those do not exist in the universe of AD! :-) –  Asaf Karagila Feb 24 '12 at 14:27
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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

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