Countable and Uncountable subsets of $\mathcal{P(\mathbb{Z})}$

Question

Suppose I have a relation on $\mathcal{P(\mathbb{Z})}$ which says give me all ordered tuples $(S,T)\in\mathcal{P(\mathbb{Z})} \times \mathcal{P(\mathbb{Z})}$ such that $S\underset{\text{bijection}}{\longrightarrow}T$. Now this is obviously an equivalence relation and then consider now the equivalence classes of $[{0}]$ and $[\mathbb{Z}]$. The question is that does $[\mathbb{Z}]$ have the same cardinality as $\mathbb{Z}$ my intution says no. $[\mathbb{Z}]$ is a infinite set of infinitely countable subsets of $\mathbb{Z}$. But the set itself $[\mathbb{Z}]$ seems to be an infinte subset of the $\mathcal{P(\mathbb{Z})}$ which we know is uncountable. On the other hand of inquiries, is the $[\{0\}]$ numerically equivalent to the set $[\{0,1,2,3,...k\}]$ where $k\in\mathbb{N}$, clearly they have the same cardinality since the class $[\{0,1,2,3,...k\}]$ contains all sets which have cardinality $\vert k+1\vert$. How would I go about proving that?

Answer 1

Consider $A=2\mathbb{Z}$, i.e. the subset of all numbers divisible by $2$ and its complement $B$. Both are countable. Moreover for any subset $C\subseteq B$ we have that $A\cup C$ is countable. And then the function $C\mapsto A\cup C$, as a function $P(B)\to P(\mathbb{Z})$ is injective. Meaning there is uncountably many countable subsets of $\mathbb{Z}$.

On the other hand there is countably many finite subsets of $\mathbb{Z}$. That's because every finite subset is a subset of some $\{-n,\ldots,-1,0,1,\ldots,n\}$ and each such subset has finitely many subsets itself. This easily implies that each $[X]$ is countable for finite $X$.