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simont

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[personal profile] jack Mon 2014-03-10 13:14
I now actually read this proof, and it sounds right to me, though I don't guarantee I haven't missed some exception.

What confused me in the pub is, you use the axiom of choice to get a basis for R over Q. But don't you just need a basis over Q for the vector space generated by the fragment lengths you actually have, not over any possible irrationals? Isn't that just some subset of the fragment lengths, throwing away any non-linearly-independent ones?
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