0
$\begingroup$

This is what I have so far.

Let U ⊆ ℝ where U is open. By a theorem we have that all real numbers are uncountable and that rational numbers are dense in ℝ.

$\endgroup$
  • $\begingroup$ All real numbers are countable??? $\endgroup$ – mathcounterexamples.net Sep 22 '16 at 19:55
  • 4
    $\begingroup$ Real numbers are uncountable. Also, there are plenty of answers online, for example here $\endgroup$ – Pawel Sep 22 '16 at 19:58
2
$\begingroup$

U=$\cup_{A}(ri,rj)$ where A is the set of rational tuples, so that (ri,rj) is a subset of U. Obviously this is a subset of U. Now you habe to show, that it contains U: take any element of U: u, than because u is open, there is an open ball B with u in B. This ball can be arbitraily made smaller, so that it has rational borders. So u is in an open interval with rational borders, therefore in the Union. The Union is countable, because the Rationals and therfore $\mathbb{Q}^{2}$ are.

$\endgroup$
  • $\begingroup$ Wait, why do you consider tuples? This was for $\mathbb{R}$ and not $\mathbb{R}^2$? Or are you referring to end points? $\endgroup$ – mavavilj Sep 19 '18 at 20:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.