Is the cardinality of the rationals the same on the unit interval? Let $\mathscr{G}$ be a compact subset of $\mathbb{R}$. Denote by $\mathbb{Q} ( \mathscr{G})$ the set of all rational numbers in $\mathscr{G}$. Does $| \mathbb{Q} ( \mathscr{G}_1)| = |\mathbb{Q} (\mathscr{G}_2)|$ for $\mu(\mathscr{G}_1) > \mu(\mathscr{G}_2)$ where $\mu$ is the Lebesgue Measure?   
i.e. are there more rationals in $[0,2]$ than $[0,1]$?

Assuming that $a<b$, and $c<d$ are all rational, one can set up 1-1 correspondence between rational numbers in $[a,b]$ with those in $[c,d]$ by the mapping
$$
y = \frac{d-c}{b-a}(x-a) + c, ~~ x \in [a,b], ~~y \in [c,d] \tag1$$
Added in response to OP's question
Note that (1) is invertible:
$$
x = \frac{b-a}{d-c}(y-c) + a \tag2
$$
Thus for every $x\in[0,1]$ we can find a $y \in [c,d]$ and vice-versa. Note that if $x$ is irrational then so is $y$ and vice-versa. So the irrationals can also be put in to a 1-1 correspondence. The fact that one is countable infinity and the other uncountable is incidental. The key point is that there is a 1-1 correspondence.
