I posted a question on . The answer is brilliant and very illuminating.
The statement of the problem is:
Let us consider , putting all the elements in the set in a sequence, denoted . We define . Notice that .
So we define . We have , and is a zero-measure set.
So how to find an irrational number in ?
Robert Israel gives a very clever solution:
For an “explicit” construction, consider , where given , if . Note that $a_{m(n)} <z<a_{m(n)} +2^{ 1−k_{n+1}} <a_{m(n)} +2^{−2m(n)} so . Since , is in all the . Its base-2 expansion contains infinitely many 1’s but also has arbitrarily large gaps between the 1’s, so can’t be eventually periodic, and thus must be irrational.