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Let's consider the real numbers as a $\mathbb{Q}$-linear space and denote it by $\mathbb{R}_{\mathbb{Q}}$. It was proved here that $\dim \mathbb{R}_{\mathbb{Q}} > \aleph_0$. I suspect that the dimension is actually $\mathfrak c$. But how could we prove it's really $\mathfrak c$ ?

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marmistrz
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    See http://math.stackexchange.com/q/67751 – Watson Feb 17 '16 at 13:47
  • @Watson Which answer there are you talking about? – David Feb 17 '16 at 13:49
  • If $\kappa$ is an infinite cardinal then a set of cardinality $\kappa$ has only $\kappa$ finite subsets... – David C. Ullrich Feb 17 '16 at 13:49
  • @David : I edited the link. It is Leon's answer, for instance. – Watson Feb 17 '16 at 13:49
  • @The answer by Leon in the purported duplicate seems unnecessarily complicated, and the other answers answer a different question than this one. – David Feb 17 '16 at 13:52
  • @David : I agree. But Arturo Magidin added a simple argument in his answer. – Watson Feb 17 '16 at 13:55
  • @Watson He states that the dimension is uncountable, not that it is $2^{\aleph_0}$. – David Feb 17 '16 at 14:01
  • @David : is my following argument correct? A $\Bbb Q$-basis $B \subset \Bbb R$ has cardinality at most $2^{\aleph_0} = |\Bbb R|$. Since $B$ is not countable, we get $|B|>\aleph_0$. Assuming the continuum hypothesis, we have $|B|=2^{\aleph_0}$. Is it necessary to use to continuum hypothesis? – Alphonse Mar 05 '16 at 21:21
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    @Alphonse The argument is correct if you assume the continuum hypothesis. But that's unnecessary. A vector space over a countably infinite field has the same cardinality as its basis. – David Mar 05 '16 at 23:13
  • @David : yes, I see. If $V$ is an infinite-dimensional vector space over a countably infinite field $k$ then $V \cong k^{(J)}$ where $J$ is an infinite set. Then $$|V|=|k^{(J)}|=\left| \bigsqcup\limits_{I \in \mathcal P} k^I \right| = |\mathcal P| \cdot |k| = |\mathcal P| = |J|$$ where $\mathcal P$ denotes the collection of finite subsets of $J$. – Alphonse Mar 06 '16 at 17:24
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    @Alphonse Right, although in your argument some of those inequalities should at first be written as inequalities. Then apply Cantor-Bernstein. – David Mar 06 '16 at 19:14
  • Related: http://math.stackexchange.com/questions/583601 – Alphonse Aug 17 '16 at 12:14

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