What is the cardinality of the set of all functions from $\mathbb{Z} \to \mathbb{Z}$?

Question

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?

Answer 1

$ 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 $F\in B'$ : $$F(1)=1,\ F(i)=0 \ (2\leq i \leq 1+ f(1)),\ \cdots $$

Answer 2

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 $2^\omega=\mathbb R$.

Answer 3

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: $$\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \left|(2^{\mathbb{Z}})^{\mathbb{Z}}\right| = 2^{|\mathbb{Z} \times \mathbb{Z}|}$$