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A natural vector space is the set of continuous functions on $\mathbb{R}$. Is there a nice basis for this vector space? Or is this one of those situations where we're guaranteed a basis by invoking the Axiom of Choice, but are left rather unsatisfied?

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This is what Mick Jagger was singing about. –  copper.hat Apr 25 '12 at 6:04
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Cue the AC guys... Oh wait, that's me. Let me write something up. :-) –  Asaf Karagila Apr 25 '12 at 7:43
    
Hmmm. Since I don't see how the continuous functions make a Polish group, I can't really use the "usual" tools for this sort of proof. I think it is very unlikely that a Hamel basis exists without the aid of the axiom of choice here. If I can come up with a clean argument I'll post it later today. –  Asaf Karagila Apr 25 '12 at 9:10
    
@Asaf Karagila: Some papers by Lorenz Halbeisen at iam.unibe.ch/~halbeis/publications/publications.html may be relevant (perhaps not directly, but the references might be helpful), such as #5, 9, 21. –  Dave L. Renfro Apr 25 '12 at 15:35
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@AsafKaragila: How about if we equip $C(\mathbb{R})$ with the topology of uniform convergence on compact sets. This gives us a separable Frechet space and in particular a Polish vector space. Now can you tell us about your "usual" tools? –  Nate Eldredge May 29 '12 at 12:47
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There is, in a fairly strong sense, no reasonable basis of this space. Zoom in on a neighborhood at any point and note that a finite linear combination of functions which have various kinds of nice behavior in that neighborhood also has that nice behavior in that neighborhood (differentiable, $C^k$, smooth, etc.). So any basis necessarily contains, for every such neighborhood, a function which does not behave nicely in that neighborhood. More generally, but roughly speaking, a basis needs to have functions which are at least as pathological as the most pathological continuous functions.

(Hamel / algebraic) bases of most infinite-dimensional vector spaces simply are not useful. In applications, the various topologies you could put on such a thing matter a lot and the notion of a Schauder basis becomes more useful.

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That is probably the best explanation I've seen as to why not to usel Hamel bases in real/complex function spaces. +1! –  Patrick Da Silva May 29 '12 at 22:41
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Using Nate Eldredge's comment we have that $C(\mathbb R)$ is a Polish vector space.

Consider a Solovay model, that is ZF+DC+"All sets have the Baire property". In such model all linear maps into separable vector spaces are continuous, this is a consequence of [1, Th. 9.10].

It is important to remark that a continuous function (from $\mathbb R$ to $\mathbb R$) from a compact set is uniformly continuous is a result which do not require any form of choice, and I believe that Dependent Choice (DC) ensures that uniform converges on compact sets is well behaved.

Suppose that there was a Hamel basis, $B$, it has to be of cardinality $\frak c$. So it has $2^\frak c$ many permutations, which induce $2^\frak c$ different linear automorphisms.

However every linear automorphism is automatically continuous, so it is determined completely by the countable dense set, and therefore there can only be $\frak c$ many linear automorphisms which is a contradiction to Cantor's theorem since $\mathfrak c\neq 2^\frak c$.

This is essentially the same argument as I used in this answer.


Bibliography:

  1. Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.
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Very nice! So in fact one certainly needs the full axiom of choice, and DC is not enough. –  Nate Eldredge May 29 '12 at 15:56
    
@Nate: Indeed. DC itself is often not enough. Shelah even showed that DC($\aleph_1$) is not enough to have sets without the Baire property! –  Asaf Karagila May 29 '12 at 16:00
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I'll use faith to believe we are in one of those situations describe by the axiom of choice ; had one discovered a useful basis for this vector space, it'd be known all over the place. The best we have as a basis right now, (and the word "best" means 'to my belief, the one that looks the prettiest') is the fact that the functions $e^{inx}$ with $n \in \mathbb Z$ are a basis of the space $L^2([a,b])$ of $\mathbb C$-valued functions. For real valued functions, take the functions $\sin(nx)$ and $\cos(nx)$ as your basis.

Someone help me here ; I know the notion of basis I speak of here is well-defined, i.e. that we say that an infinite series represents the element of the vector space if the infinite series converges. This is clearly not an Hamel basis, where we require the linear combination to be finite. Is there a name for this kind of space? At first I thought the name Hamel was given to this infinite-convergence-notion of basis, but now I realize that the Hamel name was given to the algebraic basis.

Hope that helps,

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just a remark on your last claim - that's not literally true since you need to take convergent infinite "linear combinations" whereas in the definition for a basis for a vector space, finite linear combinations give you everything. –  user29743 Apr 25 '12 at 5:34
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Hamel basis is what you want here. –  copper.hat Apr 25 '12 at 5:36
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Patrick, do you mean $e^{inx}$? –  Isaac Solomon Apr 25 '12 at 5:36
    
Sorry for the lack of clarity in my comment ; indeed, all what you said was correct, and was what I meant, I just wrote it in a hurry with my mobile phone. Now that I am at home in front of my computer I can see that I wrote too sketchy =) Thanks for the comments. –  Patrick Da Silva Apr 25 '12 at 6:10
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The "topological" basis is called Schauder basis. –  Asaf Karagila Apr 25 '12 at 9:07
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