equivalence relation - checking

Question

At $\mathcal P(\mathbb{Z})$ we define equivalence relation $\equiv$ :

$A \equiv B \iff (A=B \vee (A \cup B)\cap\mathbb{N}=\emptyset)$

a)show that $\equiv$ is a equivalence relation at $\mathcal P({\mathbb{Z}})$

I'm not sure if it's proper but can I prove it using only one condition $A=B$ ? from it follows immediately that it's equivalence relation

b) find cardinality of equivalence classes

if $A \in \mathcal P(\mathbb{Z\setminus N})$ then $|[A]|=c$ since then from $A \equiv B \iff (A \cup B)\cap\mathbb{N}=\emptyset$ we have $B \in \mathcal P(\mathbb{Z\setminus N})$ and $|\mathcal P(\mathbb{Z\setminus N})|=c$

if $A \not\in \mathcal P(\mathbb{Z\setminus N})$ we have only $A \equiv B \iff (A=B)$ so $|[A]|=1$ since it's only one set that $A=B$

c) find cardinality of quotient set $|Q|$

we have $A \subseteq \mathcal P(\mathbb{Z})$ so $|Q| \le c$ but if $A \neq B$ then $A$ and $B$ are at different equivalence classes so since $A \in \mathcal P(\mathbb{Z})$ then $|Q| \ge c $ so $|Q|=c$

are my reasoning correct ?

Answer 1

For the first point, we have for all $A \in P(\mathbb{Z})$ $A\equiv A$ because $A=A$.

let $A, B \in P(\mathbb{Z})$ such that $A\equiv B$ so $(A=B \vee (A \cup B)\cap\mathbb{N}=\emptyset)$ and then $(B=A \vee (B \cup A)\cap\mathbb{N}=\emptyset)$ so $B\equiv A$.

Let $A, B, C \in P(\mathbb{Z})$ such that $A\equiv B$ and $B\equiv C$ we need to proof that $A \equiv C$ so we have two cases :

• If $A=B$ so $B\equiv C$ implies that $A\equiv C$.

• If $ (B \cup A)\cap\mathbb{N}=\emptyset$ and $(B \cup C)\cap\mathbb{N}=\emptyset$ so :

$$\emptyset =\emptyset\cup \emptyset =[(B \cup A)\cap\mathbb{N}] \cup [(B \cup C)\cap\mathbb{N}] =[(B \cup A\cup C)\cap\mathbb{N}] \supset (A\cup C)\cap\mathbb{N}$$ that mean $(A\cup C)\cap\mathbb{N}=\emptyset$ and so $A\equiv C$

For b) if we have $A\equiv B$ so we have two cases :

• If $ (B \cup A)\cap\mathbb{N}=\emptyset$ so $B \cap \mathbb{N} = A \cap \mathbb{N}=\emptyset$ and that mean $A,B\in P(\mathbb{Z}-\mathbb{N})$, and reciprocally if we have $A,B\in P(\mathbb{Z}-\mathbb{N})$ that implies $(B \cup A)\cap\mathbb{N}=\emptyset$. so we have our first class $P(\mathbb{Z}-\mathbb{N})$, of cardinal $c$ as you say.

• If $A\not\in P(\mathbb{Z}-\mathbb{N})$ so $A\cap \mathbb{N}\neq \emptyset$ so $A\equiv B$ if and only if $A=B$, so for each $A \not\in P(\mathbb{Z}-\mathbb{N}) $ $A=[A]$ and so cardinal of $[A]$ is one in this case.

and for the last question I described all elements of $Q$ and we see that $P(\mathbb{N})\subset Q \subset P(\mathbb{Z})$ so $|Q|= c$