Stupid question - why can't interval covers of a dense nullset be expanded to arbitrarily small covers of the real line?

Question

So we all know that the rational numbers are dense, meaning that for any positive real, however small, there's one within that distance of any given real number, and we all know that their collection has measure zero, meaning that for any positive real, there exists a countable set of intervals that contains every rational number whose lengths add up to less than that real.

So here's my question: given such a set for a given real, if you were to broaden the first interval by that real, then the next by half, the next by a fourth, the next by an eighth, and so on - since the rationals are dense, wouldn't every real number fall into one of those intervals? But you'd only be tripling the measure of the cover.

(I know this is a stupid question. I await a stupid answer.)

Answer 1

Consider a closed interval eg $[0,3]$, which is compact. Suppose the intervals surrounding the rationals cover this whole interval - this is an open cover and has a finite subcover by compactness. But if the intervals have length $\frac 12, \frac 14 \dots \frac 1{2^n} \dots$ the total length of a finite subset is less than $1$.

Try constructing a cover of the rationals which avoids $\sqrt 2$ as an exercise to see what is going on.

Answer 2

Let be $\Bbb Q = \{q_n\}_{n\in\Bbb N}$ and $\{(q_n - 1/2^n,q_n + 1/2^n)\}_{n\in\Bbb N}$ the cover.

The fact is that for some (many!) "rebel" real number $x_0$: $$\forall n\in\Bbb N: |x_0 - q_n|\ge 1/2^n.$$ Having many rational numbers arbitrarily near of $x_0$ does not mean that some concrete rational number is as near of $x_0$ as you want.