What is the cardinality of the set of all functions from $\mathbb{Z} \to \mathbb{Z}$? How can I approach this?
I have to find the cardinality of the set of the functions from $\mathbb{Z} \to \mathbb{Z}$ and I have no idea on how to solve it.
Can someone hint me here?
The approach itself is what confuses me... how do I try to map this to something else since its a set of functions?
 A: One approach is to hypothesize some cardinal $\kappa$ for which you can prove $\kappa \leq \left|\mathbb{Z}^{\mathbb{Z}}\right|$ and $\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \kappa$.
Here's a hint: 
$$\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \left|(2^{\mathbb{Z}})^{\mathbb{Z}}\right| = 2^{|\mathbb{Z} \times \mathbb{Z}|}$$
A: The given set has cardinality $\omega^\omega$. Clearly, this is greater than or equal to $2^\omega$. Conversely, it's less than  or equal to $(2^\omega)^\omega=2^{\omega^2}=2^\omega$. Thus, by CSB, the cardinality is equal to $2^\omega=\mathbb R$. 
A: $ A:=\{ f : \mathbb{Z}\rightarrow \mathbb{ Z} \} ,\ B:=\{ f: \mathbb{
    Z}\rightarrow \{ 0,1 \} \} $ so that $$ |A|\geq
   |B|= 2^{\aleph_0 }=
\mathfrak{c}$$ where $
\mathfrak{c}=|\mathbb{ R}|$ and $
    \aleph_0=|\mathbb{N}|$ (cf. Schroder-Bernstein theorem)
Since $|\mathbb{Z}|=|\mathbb{N}|$, then $|A|=|A'|$ where $A':=\{f :
\mathbb{N}\rightarrow \mathbb{N}\}$. And $|B|=|B'|$ where $B':=\{f : \mathbb{
N}\rightarrow \{0 , 1\} \}$
Now we will match an element in $A'$ into an element in $B'$.
For $f\in A'$, then define $$ 1\underbrace{0\cdots 0}_{f(1)-\text{times}}
1\underbrace{0\cdots 0}_{f(2)-\text{times}} 1 \cdots $$
so that we have $F\in B'$ : $$F(1)=1,\ F(i)=0 \ (2\leq i \leq 1+
f(1)),\ \cdots $$
