Rational number, dense but measure zero

Question

When calculating the measure of Q in real number interval [0, 1], an interval $ (q_n-\epsilon, q_n + \epsilon)$ around each rational number $ q_n $ is defined to show the measure of Q is zero.

1. Is it possible to find such interval to cover single rational number?
2. Do the rational numbers exist as if they are points in that real number interval?

However small the $ \epsilon $ is, there must be irrational numbers in $ (q_n-\epsilon, q_n + \epsilon)$ which means that there is at least one more rational number because the Q is dense.

I think I missed something, but I don't know what it is.

Answer 1

When covering the rationals bit small intervals you are right: there will be overlap, but that's not a problem. The point is that you will still get stuff less than $\epsilon$ overall, so it's still OK.

For example: cover the $n^{th}$ rational by an interval of size $\epsilon 2^{-n}$ then the total sum of all the rational lengths is

$$\sum_{n=1}^\infty \epsilon2^{-n}=\epsilon$$

so that the rationals have measure $\mu(\Bbb Q)\le \epsilon$ for every $\epsilon >0$, by the monotonicity of a measure, remember

$$\mu\left(\bigcup_{n}A_n\right) \le\sum_n\mu(A_n)$$

with equality only occasionally. Hence the rationals have measure $0$, since we can cover them by a set of as small a measure as we like. It doesn't matter if the intervals over count things, so long as they get everything, there can be some extra.