3
$\begingroup$

Let $A$ be the subset of the rationals whose denominators are prime. Can we use the fact that the rationals are dense in the reals to show $A = \mathbb{R}$? I've read that using the fact that there are infinitely many primes is useful for this but I don't see the connection.

  • 1
    But $A$ does not equal $\bf R$. Perhaps you mean the *closure* of $A$ is $\bf R$.2012-11-17

1 Answers 1