Is there a one-to-one corrospondence between integer functions and the reals?

Question

Assuming $\Bbb{Z}^\Bbb{Z}$ to mean the set of all functions from $\Bbb{Z}$ to $\Bbb{Z}$, are $\Bbb{Z}^\Bbb{Z}\cong\Bbb{R}$?
Another way to put it, assuming a set of functions $Z:=\{f:\Bbb{Z}\rightarrow\Bbb{Z}\}$, does there exist a bijective function $g:\Bbb{Z}^\Bbb{Z}\rightarrow\Bbb{R}$ that assigns a real number for every element of $Z$?
In essence, is the following statement true:
"for any function $h\in Z$, there exists one, and only one, real number $r$, such that $g(h)=r$ and $g^{-1}(r)=h$."

Answer 1

To follow GEdgar's hint, consider $r\in\mathbb{R}$. Then $r$ can be uniquely written as $$ r=\pm a.a_1a_2a_3\cdots $$ where $a$ is a positive integer and $a_k\in\{0,1,2,\cdots,9\}$ such that $a_k$ is not "eventually" $9$. This gives an injective map from $\mathbb{R}$ to $\mathbb{Z}^{Z}$.

For the other direction, consider instead $\mathbb{N}^{\mathbb{N}}$. Given any sequence of natural numbers $q_n$, it represents a "unique" real number $$ r=a_0.a_1a_2a_3\cdots $$ except for countably many sequences that are "eventually" $9$.

Now you only needs to show $\mathbb{N}^{\mathbb{N}}\cong \mathbb{Z}^{\mathbb{Z}}$.