Cardinality of two equivalence classes and quotient set

Question

In the set $\mathcal{P}(\mathbb{N})\setminus \{\emptyset\}$ we define an equivalence relation as: $$X \equiv Y \iff \left( \forall n \in X \ \exists m \in Y \ (m \ge n) \ \land \forall k \in Y \ \exists l \in X \ (l \ge k)\right)$$ I want to consider the cardinality of two equivalence classes: $[\{2\}]_\equiv$ and $[\mathbb{N}]_\equiv$, and the cardinality of the quotient set $(\mathcal{P}(\mathbb{N})\setminus \{\emptyset\})\ / \equiv$. This is a relatively new topic for me. How may I approach this problem?

Answer 1

hint

As HumbleStudent pointed out in the comment, the key is to understand the condition given for equivalence. Here is the hint:

For $X=\{2\}$, the possible $Y$'s can be only $\{1,2\}$ and $\{2\}$. The reason for that is: if $a \in Y$ is such that $a>2$, then it will violate the criterion for equivalence as there is no element in $X$ that will be greater than or equal to $a \in Y$.

This shows that $|[\{2\}]_{\equiv}|=2$.

Can you now deal with $X=\mathbb{N}$?