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We can order the rational numbers between 0 and 1 as follows:

1/1, 1/2, 1/3, 2/3, 1/4...

Now we can prove that the rationals have measure 0 as follows: For each e, construct an interval of length e/4 around the first point, e/8 around the second and so on halving each time. As this is a geometric sequence, the sum is $<e$. Since we can do this for arbitrarily small e, the set has measure 0. However the set is also dense.

Can we explicitly name a number that won't be included in these covering sets once e is less than a particular value? If not, can we produce another dense, measure 0 set which can be shown not to include a particular number.

Glorfindel
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Casebash
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  • @quasi: I edited the question to make it clearer. – Casebash Oct 17 '18 at 12:17
  • @quasi: Yes, but it is less than 1 for sensible e meaning that there are points not included. And they won't be included for any covering with an even smaller e – Casebash Oct 17 '18 at 12:41
  • @quasi OP is saying that the rationals have measure zero because they can be covered by measure $e$ sets for any $e$, not that the covers have measure zero. – guy Oct 17 '18 at 14:52
  • Yes, it seems I misunderstood. – quasi Oct 17 '18 at 15:01
  • Order your rationals as $q_1,q_2,\dots$ and consider the covering $\bigsqcup_{n \ge 0} (q_n-\varepsilon/2^n, q_n+\varepsilon/2^n) \setminus {\sqrt{2}}$. – Najib Idrissi Oct 17 '18 at 15:06
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    Seems like a duplicate of https://math.stackexchange.com/questions/151384 – usul Oct 17 '18 at 15:53

1 Answers1

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The golden ratio is more than $1/(3q^2)$ from every fraction $p/q$ so make sure that every interval around $p/q$ is narrower then that.

Empy2
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