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In the study of the vector space of the reals over the rationals, one can easily see that it has infinite dimension as $$\{\log2, \log3, \log5, \log7,\cdots\}$$ for all primes, is a infinite subset which is linearly independent, because $$\sum_{i=1}^{\infty}c_i \log p_i =0,$$ where $c_i\in\mathbb{Q}$ and $p_i$ is the $i$-th prime, implies (because one can multiply everything is this equation by a common multiple of all the $c_i$'s) $$\prod_{i=1}^{\infty}p_i^{c_i}=1 \implies c_i=0, \quad\forall i\geq1.$$ But the Axiom of Choice appears to imply that every vector space has a basis (in this case a Hamel basis) and a Hamel basis has a finite number of elements (right?).

So I must have gotten something wrong! Is it the infiniteness of primes? (Of course not! But why?)

Asaf Karagila
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Gabriel
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  • I didn't get what vector space you would consider. – reuns May 04 '16 at 02:08
  • @user1952009 A vector space where the vectors are real numbers and the scalars are rational numbers. – Gabriel May 04 '16 at 02:09
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    so you mean seing $\mathbb{R}$ as a $\mathbb{Q}$ vector space. of course $\mathbb{R}$ cannot be a $2$ dimensional $\mathbb{Q}$ vector space, nor a $k$ dimensional one, so in every cases it invalidates your proof of the finiteness of primes. hence, you are probably left with "what is $\mathbb{R}$ indeed, and what is a Hamel basis in an infinite dimensional vector space ?". and there is this discussion http://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb, or this https://drexel28.wordpress.com/2010/10/22/the-dimension-of-R-over-Q/ – reuns May 04 '16 at 02:13
  • In fact, this vector space is really infinite-dimensional. I just knew little about Hamel basis. – Gabriel May 04 '16 at 02:18
  • yes but it is more than countably infinite-dimensional, that's the problem. in fact, it is quite the same as the space of rationals sequences, and an Hamel basis of the space of rationals sequences cannot be countable (that's why those objects are awful and need the axiom of choice for at least existing) – reuns May 04 '16 at 02:20
  • Cool! Thanks for the answer! – Gabriel May 04 '16 at 02:20
  • Linear independence of a set $S$ of vectors means that any FINITE sum $\sum_{j=1}^n c_j v_j ,$( when $v_1,...,v_n$ are $n$ distinct members of $S,$ and $n$ is any member of $N$), is $0$ unless $c_1=...=c_n=0.$ – DanielWainfleet May 04 '16 at 06:27

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and a Hamel basis has a finite number of elements (right?).

A Hamel basis can have an arbitrarily large number of elements, as many as the dimension of the space.

neth
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  • for the last sentence, note that https://en.wikipedia.org/wiki/Basis_(linear_algebra)#Analysis says "If X is an infinite-dimensional normed vector space which is complete (i.e. X is a Banach space), then any Hamel basis of X is necessarily uncountable. This is a consequence of the Baire category theorem." – reuns May 04 '16 at 02:05