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Prove $\exists$ infinite set $S \subset \mathbb{R}$ s.t. $\forall x \in \mathbb{R}$, $x$ can be represented as a unique linear combination of numbers in $S$ with rational coefficients.

I'm not asking if every vector space has a basis. I already have that proven in my notes so this is NOT a duplicate.

I have that $\mathbb{R}$ as a vector space over $\mathbb{Q}$ has a basis and therefore by definition of a basis, $\mathbb{R}$ can be represented as a linear combination of some vectors $x_i \in B_i$ multiplied by coefficients in $\mathbb{Q}$, but I don't have that this representation is unique.

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  • $\begingroup$ @DietrichBurde yeah my mistake $\endgroup$ Mar 30, 2018 at 15:52

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Since every vector space has a basis, we can let $S$ be a Hamel basis of $\mathbb{R}$ as $\mathbb{Q}$-vector space. Then every real $x$ can be represented as a unique linear combination of elements of $S$ with rational coefficients.

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  • $\begingroup$ I think I'm supposed to show that that exists though. I don't know how $\endgroup$ Mar 30, 2018 at 16:05
  • $\begingroup$ This has been proved at this site. It follows from the axiom of choice $\endgroup$ Mar 30, 2018 at 16:05
  • $\begingroup$ You mean "Every vector space has a vector space basis"? $\endgroup$ Mar 30, 2018 at 16:07
  • $\begingroup$ Yes, see this duplicate. $\endgroup$ Mar 30, 2018 at 16:08
  • $\begingroup$ That question is similar, but it still doesn't give uniqueness? $\endgroup$ Mar 30, 2018 at 16:47

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