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Consider the vector space $\mathbb{R}$ over $\mathbb{Q}$. Does there exist a countable set $S \subseteq \mathbb{R}$ such that $L(S) = \mathbb{R}$?

My answer: No, there doesn't exist such a set $S$.

I'm struggling with a rigorous proof, though. Here's what I think (please help me fill in the gaps):

If $L(S) = \mathbb{R}$, then $\forall x \in \mathbb{R}$ $\exists \alpha_i \in \mathbb{Q},x_i \in S$ such that $x = \alpha_1x_1 + \alpha_2x_2 ... + \alpha_nx_n$ for some $n \in \mathbb{N}$.

Isn't this as good as a map $f:S \to \mathbb{R}$? The existence of such a map is not possible since $S$ is countable, and $\mathbb{R}$ is not. Please help me formalize this though, if it's correct. I'm unsure how to take it from here.

stoic-santiago
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    start by showing $L(S)$ is countable. – David C. Ullrich Sep 26 '20 at 11:02
  • Yes, it is possible. As given in the answer @weierstrash, if $S$ has countably infinite non-zero elements, then it could span $\mathbb{R}$ – MAS Sep 26 '20 at 13:33
  • @DavidC.Ullrich Could you share some more insight into how I can do that? – stoic-santiago Sep 26 '20 at 15:15
  • @MabudAli No, it's not possible for a countable set to span $\Bbb R$. I looked, and can't imagine what part of the answer sounds to you like it's saying this is possible. – David C. Ullrich Sep 26 '20 at 17:18
  • For each $n$, say $L(S)_n$ is the subset of $L(S)$ consisting of the elements that are linear combinations with exactly $n$ terms. Since a countable union of countable sets is countable you only need to show $L(S)_n$ is countable. But there's an obvious map from $\Bbb Q^n\times S^n$ onto $L(S)_n$. – David C. Ullrich Sep 26 '20 at 17:21
  • @DavidC.Ullrich, Yes, my comment is wrong. A countable set can never span a uncountable set. Actually, I had in mind another thing that fraction field of $\mathbb{Q}$ (using the elements of $\mathbb{Q}$ is just $\mathbb{R}$. But it is different thing I realized now. – MAS Sep 27 '20 at 04:02
  • @DavidC.Ullrich, Yes, my comment is wrong. A countable set can never span a uncountable set. Actually, I had in mind another thing that fraction field of $\mathbb{Q}$ (using the elements of $\mathbb{Q}$ is just $\mathbb{R}$. But it is different thing I realized now. – MAS Sep 27 '20 at 04:02
  • @strawberry-sunshine, The comment by David is all about.However there is another answer here. – MAS Sep 27 '20 at 04:06

1 Answers1

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HINT: If $V$ is a countable dimensional vector space over a field $k$, then $V$ is isomorphic to the countable product $k^{\omega}$.

General Grievous
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