Baire Category theorem - Determining First or Second Category I don't have a mathematician background (I am an engineer). I understand some concepts but the Baire Category theorem and the ideas of first and second category are still very abstract for me.
I have to show the following:

1. Of what category is the set of all rational numbers: 
a. in $\mathbb{R}$?
b. in itself (taken with the usual metric)?
2. Of what category is the set of all integers: 
a. in $\mathbb{R}$?
b. in itself (taken with the metric induced from $\mathbb{R}$)?

In both of the problems (b), why does the metric affect the problem?
Thanks for your time!
 A: A set is of first category if it can be written as a countable union of sets whose closure has empty interior. (A set whose closure has empty interior is commonly called nowhere dense, but I won't use this term in the remainder of this answer.) Otherwise it is of second category. If you aren't already familiar, the closure of a set $A$ is the smallest closed set which contains $A$; the interior of $A$ is the largest open set which is contained in $A$. If you're not familiar with, for instance, how you can think of $\mathbb{Z}$ as a subset of $\mathbb{R}$ vs. $\mathbb{Z}$ as a subset of itself with the topology inherited from $\mathbb{R}$, I would suggest you look up the term "subspace topology".
Let's look at the part about the integers. Purely in set theory (no topology here), we can write $\mathbb{Z} = \bigcup_{n=-\infty}^\infty \{ n \}$, which is a countable union. 
In the topology of $\mathbb{R}$, the closure of $\{ n \}$ is $\{ n \}$ and the interior of $\{ n \}$ is $\emptyset$ (because $\{ n \}$ does not contain any intervals). So we've shown that $\mathbb{Z}$ is of first category as a subset of $\mathbb{R}$. 
In the topology of $\mathbb{Z}$, the closure of $\{ n \}$ is still $\{ n \}$. However, in the topology of $\mathbb{Z}$, sets with only one element are actually open as well as closed. You can see this because $\{ n \}=\{ m \in \mathbb{Z} : |m-n|<1/2 \}$, which is the intersection of an open interval with $\mathbb{Z}$. This means that the interior of $\{ n \}$ is $\{ n \}$. That is, we have written $\mathbb{Z}$ as a countable union of sets, but the closures of these sets do not have empty interior. The Baire category theorem says that in fact you cannot write $\mathbb{Z}$ as a countable union of sets whose closure has empty interior, because $\mathbb{Z}$ is a complete metric space.
The situation is different for $\mathbb{Q}$, which is not a complete metric space (its completion is $\mathbb{R}$). As a result the answer to 1b turns out to be different than the answer to 2b.
