0
$\begingroup$

Is it hard or proven to be impossible to construct basis $B$ of $\mathbb{R}$ over $\mathbb{Q}$?

Small question regarding the cardinality: (If some miracle happened and CH turned out to be false, then cardinality of $B$ would have the possibility to be strictly between cardinality of natural numbers and real numbers. ) I know by Countable/uncountable basis of vector space that it can't have countable basis.

$\endgroup$
1
  • 4
    $\begingroup$ The cardinality is that of $\mathbb{R}$. You must invoke at least weak choice to prove its existence, so there is no possible "explicit" basis/ "construction", as any basis yields a non-Lebesgue measurable subset of $\mathbb{R}$, and Solovay constructed models of ZF without Choice in which every subset of $\mathbb{R}$ is measurable. $\endgroup$ Commented Jun 1 at 21:54

1 Answer 1

1
$\begingroup$

If we assume that $|B|<2^{\aleph_0}$ we would have $$|\mathbb{R}| = |\mathbb{Q}|\cdot|B| < \aleph_0 \cdot 2^{\aleph_0} = 2^{\aleph_0}$$ which is not true even when CH is false. We will always have $\dim_\mathbb{Q}(\mathbb{R})=2^{\aleph_0}$, CH does not affect this.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .