Trying to do an easy proof about countable sets. I'd like to prove that every time $\mathbb{Z}$ appears can be changed by $\mathbb{N}$. Seems intuitive enough for me, but I can't find a formal way to prove it.
Using $\mathbb{N}\sim\mathbb{Z}$ I proved particular cases separately, but I'd like to find a way to prove that this is always true without need to sketch a new proof for every new case (which includes $[0,1]^{\mathbb{Z}}, \mathbb{N}^{\mathbb{N}}$, and all kind of possible cases). Also I'm assuming that it is true since as I said, seems intuitive, but is it?, can $\mathbb{Z}$ be changed by $\mathbb{N}$ no matter the case?.
 A: If you use $\mathbb{Z}$ only in context of size, i.e. $|\mathbb{Z}|$ then yes, you can swap it with $\mathbb{N}$, because $$|\mathbb{Z}| = \aleph_0 = |\mathbb{N}|.$$
However, generally, you cannot swap it. If the simple example $(2-3) \in \mathbb{Z}$ does not convince you, consider the set of strictly decreasing sequences:
\begin{align} A_\mathbb{Z} &= \Big\{ (a_i)_{i \in \mathbb{N}} \in \mathbb{Z}^\mathbb{N}\ \Big|\ 
(a_i)_{i \in \mathbb{N}} \text{ is a strictly decreasing sequence} \Big\}\\
&= \Big\{ (a_i)_{i \in \mathbb{N}} \in \mathbb{Z}^\mathbb{N}\ \Big|\ 
\forall i \in \mathbb{N}.\ a_{i+1} < a_i \Big\}.
\end{align}
Observe that $|A_\mathbb{Z}| = 2^{\aleph_0} = \mathfrak{C}$, but for analogous definition of $A_\mathbb{N}$ we would have $|A_\mathbb{N}| = 0$, that is, an uncountable set becomes after the change empty (i.e. countable).
I hope this helps $\ddot\smile$
A: Obviously you can't change $\mathbb{Z}$ for $\mathbb{N}$ all the time; consider the statement $$\exists x\in \mathbb{Z}: x+2=0$$
When you reflect a bit more about the situations when you can swap the two, you will be closer to your answer, which is a bijection between the two sets.
