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

Thanks.

Kishalay Sarkar
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Aspirin
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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. – Sangchul Lee Feb 24 '12 at 13:45
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    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 Answers1

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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.

Asaf Karagila
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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
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    @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
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    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
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    @Willie: Indeed, all those do not exist in the universe of AD! :-) – Asaf Karagila Feb 24 '12 at 14:27
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    I agree with you of course. (We are crossing the beams! This conversation is getting strange because of it.) That particular passage is implicitly describing the state from the point of view within ZFC. And I am willing to bet that your comment is precisely how he has in mind of proving that those explicit constructions cannot exist. – Willie Wong Feb 24 '12 at 14:29
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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