# What is wrong in this characterisation of open sets containing all rationals?

I am having a problem understanding what's wrong with this:I know statement 1 is false and 2 is true but I can't find the mistake. 1)Any open set in $R$ is a countable union of disjoint open intervals. So if we take any open set containing all rationals its complement contains only irrational points. But because endpoint of one interval is the starting point of the next, we can associate the right endpoint of any interval(which is irrational) to each interval. Hence the number of irrational points in the complement is countable. 2)We can construct an open set in $R$ with measure less than $1$ containing all rationals. Hence its complement will have infinite measure and is uncountable.

• Presumably by "measure less than $1$" you mean "finite measure," since $\mathbb{R}$ has measure $\infty$, not $1$ ... Commented Oct 23, 2017 at 5:30
• I was talking about the open set containing all rationals to have measure less than 1 Commented Oct 23, 2017 at 5:34

For example, consider the set of open intervals $$\{(0, 3), (3, 3.1), (3.1, 3.14), (3.14, 3.141), ..., (3.142, 3.15), (3.15,3.2), (3.2, 4), (4, 5)\}.$$ It should be clear that these intervals "converge to $\pi$" - so $\pi$ isn't the right endpoint of any one of these intervals.