Separable set, the real [0,1] interval and measure I am having a hard time understanding exactly what "separable" means, and I am trying to relate that to the measure of the real [0,1] segment (which is 1, right?).
My confusion started when studying measure for Lebesgue integration. The set of rationals in [0,1] is given measure 0, so that we can integrate the indicator function of the rationals on [0,1] (which has value 0 - so the complement, the the integral of the indicator of the non-rationals on [0,1] must be 1?). 
So far so good, it's not too hard to accept that the rationals are countable, so they "count for nothing" on [0,1], which has "many more" elements. I saw a rigorous presentation where an explicit cover of the rationals in [0,1] is built and its measure brought to zero when taking a limit. No problem. 
However, I started thinking that maybe I could similarly cover the non-rationals, and bring that cover's size to 0 too. Which must fail somewhere, because that set (of non-rationals in [0,1]) must have size 1. Then I met the concept of "separability", and I started thinking: if the reals are separated by the rationals (the rationals are "dense" in $\mathbb{R}$, I can't find "two reals that are adjacent", and therefore I would be able to find a cover who limit size is 0? 
Obviously, I am making several mistakes along the way, and I am not having my topology straight. What does "separable" really entail for the reals in [0,1]? How/why will the cover size for the non-rationals in [0,1] converge to 1 rather than 0?
 A: Being separable means that there is a dense countable subset. This means that there is a subset we can write as $\{x_n\mid n\in\Bbb N\}$, such that whenever $x$ is a point in the space, there is a sequence $x_{n_k}\to x$.
That what it means being separable. Indeed $\Bbb Q\cap[0,1]$ is a dense countable subset of $[0,1]$, so it witnesses the separability.
And indeed as you said, one can enumerate the countable subset, then for every $\varepsilon$ take the interval of length $\frac1{2^n}\cdot\varepsilon$ around $x_n$, and that would amount to a cover of the countable set of measure at most $\varepsilon$.
Why would that fail if we take the irrational numbers instead?

*

*There are uncountably many irrational numbers.
This means that we can't enumerate them and use the trick of $\frac1{2^n}$. At some point we would have to resort to infinite powers of $2$, which is not a well-defined notion in the real numbers.


*Measure is volume, whereas density is an order related property (in the real numbers, anyway, as the topology is induced by the order). So while the rational numbers are dense, their volume is zero. The irrational numbers, on the other hand have a volume of $1$ (in the unit interval, of course).
This is exactly the meaning of the fact that the irrational have a full measure.


*Finally, given $\varepsilon$, we construct the $\varepsilon$-cover of the countable set, we may want to claim that it actually covers all the interval. But that would be false. Depending on the enumeration (the choice of how we indexed the $x_n$'s to begin with) we can construct irrational numbers which are not covered.
The process is similar to the construction of the Cantor set. Take the complements of all the intervals, and show that not finitely many of them have an empty intersection (which is simple enough to see). Then by the fact that these are all compact sets, the intersection is non-empty and must contain an irrational number not covered.
