Open Dense Set of Real Numbers I recently came across the set
$O=\bigcup_{n=1}^\infty O_n$ ,where
$$\mathcal{O}_n = \left(r_n - \frac{\epsilon}{2^{n + 2}}, r_n + \frac{\epsilon}{2^{n + 2}}\right)$$
where {$r_n$} is an enumeration of rationals.The above subset is dense in R and of finite measure. But my question is :
Can we explicitly construct an irrational not in the above set ?
 A: Let $O=\bigcup_{n=1}^\infty O_n$. If by 'construct' you mean that you can find a computable irrational $q$ such that $q\not\in O$, the answer is interestingly no.
Let $r_n$ represent your sequence of rationals. And let $q$ be an irrational not in $O$. If $q$ were computable there would be a subsequence $r_{n_k}$ of your sequence which converges to $q$ and such that $|r_{n_k}-r_{n_{k+1}}|<\varepsilon 2^{-(n_k+2)}$ for each $k$ (we can assume the subsequence is increasing also). That being said we get that $|r_{n_k}-q|=\sum_{i=k}^\infty |r_{n_i}-r_{n_{i+1}}|<\sum_{i=k}^\infty\varepsilon 2^{-(n_i+2)}=\varepsilon\sum_{i=k}^\infty 2^{-(n_i+2)}<\varepsilon 2^{-(n_k+2)}\sum_{i=1}^\infty 2^{-n_i}<\varepsilon 2^{-(n_k+2)}$.
Thus $q\in(r_{n_k}-\varepsilon 2^{-(n_k+2)}, r_{n_k}+\varepsilon 2^{-(n_k+2)})\subseteq O$.
Thus any such irrational not in your set is non-computable. Since computers can't even compute all computable numbers, let alone noncomputable ones, 'constructing' such an irrational seems out of the question.
EDIT
The above is incorrect. Not only is there problems with the epsilon-ics, but the statement is also false. You can certainly choose any computable irrational and then find an enumeration of the rationals such that your set avoids that irrational (for any $\varepsilon>0$ in fact).
However, if we can demonstrate an $\varepsilon>0$ and an enumeration of the rationals such that your set contains all the computable numbers, we will be able to conclude that, in general, the answer is no.
That we can do this is fairly clear. Since the computable numbers are countable, we can well order them $\{c_n\}$. And now for each $n$ we just need to pick a rational in $(c_n-\varepsilon 2^{-(n+2)}, c_n+\varepsilon 2^{-(n+2)})$. Since the rationals are computable, we can just choose that rational itself when we come to it. This sequence we construct will not be a strict bijection, but after we construct it we can go through and skip those things we have already chosen. This new sequence will not lack any computable numbers in the union either as the radii only increase. So, in general, you cannot construct an irrational not in your set.
