Why this set is of the second category?

Question

I'm watching Baire space on en.wikipedia.org, and find this conclusion.

Here is an example of a set of second category in $\mathbb R$ with Lebesgue measure zero.

$$\bigcap_{m=1}^\infty \bigcup_{n=1}^\infty \left(r_n - \frac1{2^{n+m}}, r_n + \frac1{2^{n+m}}\right)$$

where $\{r_n\}$ is a sequence that enumerates the rational numbers. my question is :

Why this set is of the second category?

Thanks a lot.

Answer 1

Every set $G_m=\bigcup_{n=1}^\infty \left(r_n - \frac1{2^{n+m}}, r_n + \frac1{2^{n+m}}\right)$ is a dense open set because is union of open sets and contains the rational numbers. So it's complement is nowhere dense, and hence $B=\bigcup_{m=1}^\infty G_m'$ is of first category which $G_m'$ is the complement of $G_m$.
Now observe that $A=B'=\bigcap_{m=1}^\infty \bigcup_{n=1}^\infty \left(r_n - \frac1{2^{n+m}}, r_n + \frac1{2^{n+m}}\right)$ must be of second category.

Note. We used De Morgan's laws in the last line.