In $\mathcal{P}(\mathbb{Z})$, $A \equiv B \iff \exists_{r \in \mathbb{Z}} ( \forall_{x \in A} (x + r \in B) ∧ \forall_{y \in B} (y - r \in A))$... We define an equivalence relation $≡$ in the set $\mathcal{P}(\mathbb{Z})$ as:
$$A \equiv B \iff \exists_{r \in \mathbb{Z}} ( \forall_{x \in A} (x + r \in B) ∧ \forall_{y \in B} (y - r \in A))$$
Determine $|\mathcal{P}(\mathbb{Z})/_≡|$ and $∣[\mathbb{N}]_≡∣ $.

I know that for $r = 0$: $A = B$, for $r = 1$: $A$ and $B$ lay next to each other, have $1$ common element and both cover a part of $X$ axis that is $1$ wide. Similar facts are true for $r = 2$, $r =3$...
That would mean that: $|\mathcal{P}(\mathbb{Z})/≡| = |\mathbb{N}|$.
However, I don't know what to do with $∣[\mathbb{N}]_≡∣ $.
 A: HINT: It is not necessarily true that $A$ and $A+1$ are one unit wide and lie next to each other: $A$ could be all of $\Bbb Z$, in which case $A+1=A$. (And in fact this is the crucial observation that lets you show that $\left|[\Bbb N]_\equiv\right|=|\Bbb N|$.)
Nor do they necessarily have one element in common. In that example they are identical and have infinitely many elements in common, and if $A=\{0\}$, $A+1=\{1\}$, and they have to element in common. Finally, no subset of $\Bbb Z$ covers a one-unit segment of the $x$-axis.
In fact the quotient has cardinalities larger than $|\Bbb N|$. I’ll point you towards one way to answer that question.
Let $\Sigma$ be the set of strictly increasing sequences in $\Bbb N$ whose first term is $1$. If $\sigma=\langle n_k:k\in\Bbb N\rangle\in\Sigma$, let $$A_\sigma=\left\{\sum_{i=0}^kn_k:k\in\Bbb N\right\}\,.$$ Show that if $\sigma,\tau\in\Sigma$, and $\sigma\ne\tau$, then $A_\sigma\not\equiv A_\tau$. Conclude that
$$\left|\wp(\Bbb N)/\!\sim\right|=\left|\wp(\Bbb Z)/\!\sim\right|\ge|\Sigma|\,.$$
Then show that
$$|\Sigma|=|\wp(\Bbb N)|=|\Bbb R|$$
and deduce that $\left|\wp(\Bbb N)/\!\sim\right|=\left|\wp(\Bbb Z)/\!\sim\right|=|\Bbb R|$.
A: Since the quotient map $\mathcal{P}(\mathbb{Z}) \to \mathcal{P}(\mathbb{Z})/\equiv$ given by $A \mapsto [A]_\equiv$ is surjective,  $$\left|\mathcal{P}(\mathbb{Z})/\equiv\right| \leq |\mathcal{P}(\mathbb{Z})| = |\mathbb{R}|.$$
To show $|\mathcal{P}(\mathbb{Z})/\equiv| \geq |\mathbb{R}|$, consider the map $\mathcal{P}(\mathbb{Z} \cap [1,\infty)) \to \mathcal{P}(\mathbb{Z})/\equiv$ given by $A \mapsto [\{0\} \cup A]_\equiv$.  Show that this map is injective.
For example, $[\mathbb{N}]_\equiv = \{\mathbb{Z} \cap [r,\infty) : r \in \mathbb{Z}\}$ has cardinality $|\mathbb{Z}|$ and the only representative with minimal element $0$ is $\mathbb{N}$ itself.
