A dense open set in the reals, with small size

It will be shown that in $\mathbb{R}$ , for any $\epsilon > 0$ , there is a dense open set $O$ such that $\mu^*(O) < \epsilon$ , where $\mu^*$ is the outer Lebesgue measure.

Consider the set of rationals, $\mathbb{Q} \subset \mathbb{R}$ , which is known to be countable. So, the elements of the rational numbers may be enumerated so that

$\mathbb{Q} = \{ q_1, q_2, ... \}$ . Now, fix $\epsilon >0$ and around each each $q_n, n\in \mathbb{N}$ , center an interval of length $\frac{\epsilon}{2^n}$ . That is, form the interval

$I_n = (q_n - 2^{-(n+1)}\epsilon, q_n + 2^{-(n+1)}\epsilon)$ .

Taking the union of all such intervals $I_n$ forms an open set, which will be called $O$ . It follows now that $\mathbb{Q} \subset O = \bigcup I_n$ . Now, the outer lebesgue measure is subadditive, so it follows that

$m^*(O) = m^*(\bigcup I_n) \leq \sum_{n=1}^\infty m^*(I_n) = \sum_{n=1}^\infty 2^{-n}\epsilon = \epsilon$ .

Since $\epsilon >0$ was arbitrary, the proof is complete.

NB: This is my first post. Readers who stumble here are free (and encouraged) to provide feedback and corrections. This blog is mostly going to consist of interesting math tricks and thoughts on subjects I find interesting.

A dense open set in the reals, with small size

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