Beside $\mathbb N, \; \mathbb Z$, and $\mathbb Q$, the following are frequently used subsets of $\mathbb R$ in Real Analysis:
Open interval: $(a,b)=\{x\in \mathbb R\;|\; a < x < b\}$.
Closed interval: $[a,b]=\{x\in \mathbb R\;|\; a\leq x\leq b\}$.
Half-open/ half-closed interval:
$$(a,b]=\{x\in \mathbb R\;|\; a < x\leq b\}, \text{\;and}$$
$$[a,b)=\{x\in \mathbb R\;|\; a\leq x < b\}.$$
Example.
The domain and range of the function $f$ defined by $f(x)=1/\sqrt{1-x^2}$ are $(-1,1)$ and $[1,\infty)$ respectively.
The preceding sets are all infinite sets. Infinite sets may be of different kinds in terms of counting their elements.
Definition (Countable set).
A set $S$ is countable if it is finite or there is a one-to-one function from $\mathbb N$ onto $S$.
A set $S$ is countable if there is a one-to-one function from $S$ onto a subset of $\mathbb N$.
An infinite countable set is called an enumerable set. By definition an enumerable set has a bijection with $\mathbb N$. Therefore a countable set is either finite or enumerable.
Example.
$\mathbb N$ is countable because of the identity map $I:\mathbb N\to \mathbb N$ defined by $I(n)=n$.
To show $\mathbb Z$ is countable, define $f:\mathbb N\to \mathbb Z$ by
$$f(n)=\begin{cases}
\frac{n}{2} & \text{if } n \text{ is even}\\
\frac{1-n}{2} & \text{if } n \text{ is odd}
\end{cases}$$
To show one-to-oneness of $f$, let $f(m)=f(n)$ for some $m,n\in \mathbb N$. Note that if both $m$ and $n$ are not even or both $m$ and $n$ are not odd, then $f(m)=f(n)\implies m+n=1$, a contradiction. So when $m$ and $n$ are even,
$$f(m)=f(n)\implies \frac{m}{2}=\frac{n}{2} \implies m=n.$$ Similarly when $m$ and $n$ are odd,
$$f(m)=f(n)\implies \frac{1-m}{2}=\frac{1-n}{2} \implies m=n.$$
To show ontoness of $f$, let $k\in \mathbb Z$. If $k>0$, then
$f(2k)=\frac{2k}{2}=k$ where $2k\in \mathbb N$. If $k\leq 0$, then $f(1-2k)=\frac{1-(1-2k)}{2}=k$ where $1-2k\in \mathbb N$.
Theorem.
$\mathbb Q$ is countable.
Since $\mathbb Z$ is countable, $S_q=\{\frac{p}{q}\;|\; p\in \mathbb Z\}$ is countable for all $q\in\mathbb N$. Note that
$$\mathbb Q=\bigcup_{q=1}^{\infty} S_q.$$
Since countable union of countable sets is countable (long exercise), $\mathbb Q$ is countable.
Theorem.
The interval $(0,1)$ is uncountable.
(Cantor's diagonal argument, 1891) Suppose $(0,1)$ is uncountable. Let $(0,1)=\{x_n \;|\; n\in \mathbb N\}$ where each $x_n$ has a decimal expansion (not necessarily finite):
$$\begin{array}{ccl}
x_1&=&0.{\bf a_{11}}a_{12}a_{13}a_{14}\cdots\\
x_2&=&0.a_{21}{\bf a_{22}}a_{23}a_{24}\cdots\\
x_3&=&0.a_{31}a_{32}{\bf a_{33}}a_{34}\cdots\\
&\vdots&\hspace{55pt} \ddots
\end{array}$$
Now construct a number $b=0.b_{1}b_{2}b_{3}b_{4}\cdots \in (0,1)$ by choosing $b_i\neq a_{ii}$ as follows:
$$b_i=\begin{cases}
6 & \text{if } a_{ii}=5\\
5 & \text{if } a_{ii}\neq 5
\end{cases}$$
Then $b\neq x_i$ for all $i\in \mathbb N$ but $b\in (0,1)$ contradicting that $(0,1)=\{x_n \;|\; n\in \mathbb N\}$.
By the preceding theorem, $(0,1)\setminus \mathbb Q$ is uncountable and so are $\mathbb R$ and $\mathbb R\setminus \mathbb Q$. Thus $\mathbb Q$ is a countably infinite set where $\mathbb R\setminus \mathbb Q$ is an uncountably infinite set.
To distinguish the sizes (cardinality) of infinite sets such as $\mathbb N$ and $\mathbb R$, Georg Cantor introduced cardinal numbers such as $\aleph_0$ (read aleph-naught or aleph-zero) $<\aleph_1<\aleph_2<\cdots$ where $\aleph_0$ is the cardinality of $\mathbb N$. The cardinality of $\mathbb R$ is denoted by $\mathfrak{c}$. By the preceding theorem, $\aleph_0<\mathfrak{c}$.
The continuum hypothesis states that $\mathfrak{c}=\aleph_1$, i.e., there is no set whose cardinality is strictly between that of $\mathbb N$ and that of $\mathbb R$.