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$\{10^z|z\in \mathbb Z\}$ looks like a basis of $\mathbb R$ over some field. However, it is definitely not a Hamel basis over $\mathbb Q$ due to

1.It only has countable elements compared with any Hamel basis actually has uncountable many.

2.Most real number needs to be represented as an infinite sum.

3.The most lethal one, itself is not independent over $\mathbb Q$.

So my question is is it another type of basis?

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    Why does it look like a basis?2012-11-16
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    "Basis" in the vector space sense is the wrong mathematical idea. You're thinking more about the fact that the set $\{0,1,2,3,4,5,6,7,8,9\}^\omega$ may be put in correspondence with real numbers, using the usual decimal representation. Without a notion of norm on the candidate basis there's not much to do, and even then, as Marc van Leeuwen stated, the field will contain $\mathbb Q$.2012-11-16
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    @AsafKaragila Because Legendre polynomials looks like an orthogonal basis on a vector space, and it has countable members. $\{10^z|z\in \mathbb Z\}$ also has countable members which are all polynomials too.2012-11-16

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No this is nothing like a basis. As you noted the elements are dependent over $\Bbb Q$, and since this is the prime field in characteristic $0$, any candidate field will contain $\Bbb Q$, and the elements given will remain linearly dependent over any such field.

If you are thinking of decimal representation of real numbers, it is not really like expression of vectors in a basis in any formal sense. Besides it has some surprises of the kind $0.99999999\ldots=1.00000000\ldots$, which never happen for vector space bases.

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    Thank you. But is Legendre polynomials an orthogonal basis on a vector space? If so, is $\{10^z|z\in \mathbb Z\}$ the same type? I'm not very clear about that...2012-11-16
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    Legendre polynomials are a (Hamel) basis of the space of polynomials. They are also a Hilbert basis of the Hilbert space $L^2([-1,1])$, which is a different notion. The set of powers of $10$ (or of any rational constant other than $1$) do not form a basis of any kind of space in any such sense, because of their linear dependency. If you want any fun in this direction at all, study the $p$-adic integers, athough there is no basis there either.2012-11-16
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    Okay, I remember. Legendre polynomials is an orthogonalization of $\{x^n|n \in \mathbb N\}$, but $x^n$ is a bit defferent from $10^n$.2012-11-17