Q(set of rationals) is of baire's first category in itself but N(natural numbers) are of second category in itself .

Question

since we need to see whether they can be written as a countable union of nowhere dense sets or not . for N , i thought {1} these single-tons are dense in N . so N is of second category. Is this reason correct? but I'm not able to find how Q is of first category in itself ..

Answer 1

The space $\mathbb N$ is of second category because its only dense subset is $\mathbb N$ itself. So, if you have a family of open dense subsets, they are all equal to $\mathbb N$.

And in $\mathbb Q$ you have$$\emptyset=\bigcap_{q\in\mathbb Q}(\mathbb Q\setminus\{q\}).$$So, in $\mathbb Q$ you can express the empty set as a countable intersection of open dense sets.

Answer 2

In $\mathbb N$ every set is closed and every set is open too. Hence the interior of the closure of $A$ equals the interior of $A$ which is $A$ itself. Hence there are no non-empty subsets that are nowhere dense. It follows that you cannot write $\mathbb N$ as a countable union of nowhere dense subsets which means $\mathbb N$ is of second category.

Since $\mathbb Q$ is the union of the sets $\{q\}, q \in \mathbb Q$ and $\{q\}$ is nowhere dense for each $q$ it follows that $\mathbb Q$ is of first category.