What do finite, infinite, countable, not countable, countably infinite mean? [duplicate]

The definitions are:

• We say that a set $A$ is finite if and only if there exists some $k\in\mathbb N$ such that there exists $f\colon A\to\{n\in\mathbb N\mid n<k\}$ such that $f$ is a bijection.

  Namely, $A$ is finite if we can write it as $\{a_0,\ldots,a_{k-1}\}$.

• We say that a set $A$ is countable if and only if there exists $f\colon A\to\mathbb N$ which is injective. Clearly every finite set is countable, but also some infinite sets are countable.

  Note that some places define countable as infinite and the above definition. In such cases we say that finite sets are "at most countable".

• We say that a set $A$ is uncountable if and only if it is not countable.

Now to show that $\mathbb{N,Z,Q}$ are countable we do the following:

1. Note that $\mathbb N$ is countable by definition (the identity function witnesses that);
2. The set $\mathbb Z$ is countable since we can consider the following function: $$f(m)=\begin{cases}2m& m\geq 0\\ -2m+1& m<0\end{cases}$$ We can show that this is an injective function (in fact a bijection) by considering the two cases.

3. To show that $\mathbb Q$ is countable we use the following facts:

   • If $A$ is countable then $A\times A$ is countable.
   • If $A$ is infinite and countable and $B$ is a set such that there is $f\colon A\to B$ which is onto $B$, then $B$ is countable, and vice versa: if $B$ is countable then such surjective function exists.

   Now we have that since $\mathbb Z$ is countable so is $\mathbb Z\times\mathbb Z$; we consider the function $f\colon\mathbb Z\times\mathbb Z\to\mathbb Q$ defined as: $$f(m,k)=\begin{cases}\frac{m}{k}&k\neq 0\\ 0 &k=0\end{cases}$$ It is not hard to see that this is a surjective function, since every rational number can be written as a fraction $\frac{m}{k}$. Therefore $\mathbb Q$ is countable too.

4. Lastly, $\mathbb R$ is uncountable. For this we use Cantor's theorem which states that if $A$ is a set then there is no surjective function from $A$ onto $\mathcal P(A)$ (the set of all subsets of $A$).

   We can use the above theorem to show that $\mathbb R$ is in fact with bijection with $\mathcal P(\mathbb N)$, and therefore $\mathbb R$ is not countable.

5. Since the above shows that $\mathbb R$ is uncountable, and $\mathbb R\subseteq\mathbb C$ we have that the complex numbers are also uncountable.