A question in set theory

Question

Is $\bigcup_{n\geq 1}\left(q_n-\frac{1}{n},q_n+\frac{1}{n}\right)=\mathbb{R}$?. Here $\{q_n\}_n$ is the enumeration of rational numbers.

Answer 1

It will depend on exactly what enumeration of the rationals is used.

• To get an enumeration where the answer is yes: since the series $\sum_{i\in\mathbb{N}}{1\over 2n}$ diverges, we can find a sequence $(p_i)_{i\in\mathbb{N}}$ of rationals such that $$\{(p_i-{1\over 2i}, p_i+{1\over 2i}): i\in\mathbb{N}\}$$ covers $\mathbb{R}$. Now let $(q_i)_{i\in\mathbb{N}}$ be any enumeration of rationals with $q_{2k}=p_k$.

• To get an enumeration where the answer is no: fix some irrational - say, $\pi$ - and "play keep-away" by enumerating all the rationals so that the $n$th rational you enumerate is at distance at least $1\over n$ from $\pi$. It's not hard to see how to do this (and in fact give a completely explicit way to turn an arbitrary enumeration of the rationals into an enumeration of this type).

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So some enumerations yield a positive answer while others yield a negative answer. One thing we can say independently of the enumeration is that there will be some sequence $(\alpha_i)_{i\in\mathbb{N}}$ tending to zero such that $\bigcup_{i\in\mathbb{N}}(p_i-\alpha_i,p_i+\alpha_i)=\mathbb{R}$; see here. So while the specific sequence $({1\over n})_{n\in\mathbb{N}}$ might not "cover everything" via your enumeration, some other similar sequence will.