Definition: A set $M\subset X$ is called "of first category" if it is countable union of nowhere dense sets. Otherwise its called "of second category".
I want to see whether the following sets are of first or second category:
1) $\mathbb Q$
2) $\mathbb R$ \ $\mathbb Q$
3) $\mathbb Z$
4) $\mathbb R$ \ $\mathbb Z$
My attempt:
1) $\mathbb Q$ is of first category because we can write it as $\mathbb Q=\bigcup_{q\in \mathbb Q}\{q\}$. This is a countable union and each $\{q\}$ is closed with empty interior.
2) $\mathbb R$ \ $\mathbb Q$ is of second category. Prove: Suppose that is is of first category, hence $\mathbb R$ \ $\mathbb Q=\bigcup_{n\in \mathbb N}A_n$ with nowhere dense sets $A_n$. Then we can write $\mathbb R=\bigcup_{n\in \mathbb N}A_n\bigcup_{q\in \mathbb Q}\{q\}$ which is a contradiction because every complete metric space if of second category.
3) Actually I thought it is of first category (Same reasoning as in 1) ). But $\mathbb Z$ is a closed subspace of $\mathbb R$, hence also complete, so it must be of second category, I guess.
4) Can be answered when I know 3)
Can someone go through it? Thanks
Questions 1 and 2 heb been proved quite correctly. The set $\mathbb{Z}$ is first category in $\mathbb{R}$, similarly as 1 is. It is however second category in itself (so the integers as a space in its own right is second category) by Baire's theorem again. $\{n\}$ is nowhere dense in the reals (as the reals have no isolated points) but not as a subset of the integers (as it is isolated, and so has non-empty interior!).
$\mathbb{Z}$ is even closed and nowhere dense, so very much first category. Its complement is again second category by the same argument, which just can be restated as "if $X$ is second catgeory in itself, and $A \subseteq X$ is first category, then $X \setminus A$ is second category in $X$", using the same proof.