Open cover rationals proper subset of R? If I were to cover each rational number by a non-empty open interval, would their union always be R? It seems correct to me intuitively, but I am quite certain it is wrong. Thanks
 A: If you enumerate the rationals as a sequence $x_1, x_2, \dots$, you can then take a sequence of open intervals $(x_1-\delta, x_1+\delta), (x_2-\delta/2, x_2+\delta/2), (x_3-\delta/4, x_3+\delta/4), \dots$ which gives an open cover for $\mathbb{Q}$ of total length $4\delta$, which can be made as small as you wish, by choosing $\delta$ sufficiently small.
A: Take an irrational number $x \in \mathbb{R} - \mathbb{Q}$. If $q \in \mathbb{Q}$ is any rational number then $\left| x - q \right| = r_q > 0$. The union $\displaystyle \bigcup_{q \in \mathbb{Q}} (q-r_q, q+r_q)$ then does not contain $x$, and so particular does not cover $\mathbb{R}$.
[Interestingly, it does cover $\mathbb{R}-\{x\}$; but you could change your intervals in the union to force this not to be the case.]
A: Take an enumeration of rational let's say $a_m$ and then cover the rationals like this 
$A_m=(a_m-\frac{1}{m^2},a_m+\frac{1}{m^2})$
Now since $ \sum \frac{1}{m^2}$ converges you cannot cover the real numbers
A: This is not true. Enumerate the rational numbers as $\{q_n\vert n\in {\bf N}\}$. 
Then consider the union $\bigcup_n(q_n-2^{-n},q_n+2^{-n})$. The set will have measure not greater than $2$, so much less than the entire real line.
A: Not necessarily. For example, the set
$$\bigcup\limits_{p,q>n} \left(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\right)$$
is a union of open balls around each rational number, yet does not contain most irrational algebraic numbers for sufficiently large $n$, by the Thue-Siegel-Roth theorem.
A: Maybe this is silly, but for a rational $q<\sqrt 2$ take the open interval to be $(-\infty,\sqrt 2)$, and for a rational $q>\sqrt 2$ take the open interval to be $(\sqrt 2,\infty)$. This would furnish a counterexample.
