0
$\begingroup$

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 $\mathfrak c$ ?

$\endgroup$

marked as duplicate by Dietrich Burde, yoknapatawpha, Dan, Silvia Ghinassi, Shailesh Feb 18 '16 at 0:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ See math.stackexchange.com/q/67751 $\endgroup$ – Watson Feb 17 '16 at 13:47
  • $\begingroup$ @Watson Which answer there are you talking about? $\endgroup$ – David Feb 17 '16 at 13:49
  • $\begingroup$ If $\kappa$ is an infinite cardinal then a set of cardinality $\kappa$ has only $\kappa$ finite subsets... $\endgroup$ – David C. Ullrich Feb 17 '16 at 13:49
  • 1
    $\begingroup$ @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. $\endgroup$ – David Mar 5 '16 at 23:13
  • 1
    $\begingroup$ @Alphonse Right, although in your argument some of those inequalities should at first be written as inequalities. Then apply Cantor-Bernstein. $\endgroup$ – David Mar 6 '16 at 19:14

Browse other questions tagged or ask your own question.