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.
Find the accumulation points of the set of rational numbers whose denominators are prime.
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real-analysis
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1But $A$ does not equal $\bf R$. Perhaps you mean the *closure* of $A$ is $\bf R$. – 2012-11-17