# Find the accumulation points of the set of rational numbers whose denominators are prime.

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.

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But $A$ does not equal $\bf R$. Perhaps you mean the closure of $A$ is $\bf R$. – Gerry Myerson Nov 17 '12 at 4:28

## 1 Answer

If there were only finitely many primes, then there would be only finitely many members of the specified set in every bounded interval, so it would have no accumulation points.

Suppose $x\in\mathbb R$. I claim $x$ is an accumulation point. To prove this, suppose $\varepsilon>0$. Find a prime $p$ so big that $1/p<\varepsilon/2$ (This is the part that we wouldn't be able to do if there were only finitely many primes). Then one member of the set $\{k/p : k\in\mathbb Z\}$ is in the neighborhood of radius $\varepsilon/2$ about $x$. If that one has $k$ a multiple of $p$, then look at its two neighbors, $(k\pm1)/p$. One of them is within $\varepsilon$ of $x$.

If instead of the set of primes we had used some set containing some composite numbers, then the proof would be more involved, since the cases where $k$ is a multiple of the denominator would not be the only cases in which the fraction $k/p$ is not in lowest terms.

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