# Cardinality of two equivalence classes and quotient set

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?

• first you have to understand when are two elements of power set (or subset of natural numbers) equivalent. If you get a hold of that, you are there. – HumbleStudent Jan 10 '17 at 22:10

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 $X=\mathbb{N}$?
• So, for $X=\mathbb{N}$, the set of possible $Y$'s is $\mathcal{P}(\mathbb{N})$ and $\lvert [\mathbb{N}]_\equiv \rvert = \mathfrak{c}$? At the same time, the quotient set would have $\lvert (\mathcal{P}(\mathbb{N})\setminus \{\emptyset\})\ / \equiv \rvert = \lvert \mathbb{N} \rvert$. Is this correct? – Zelazny Jan 11 '17 at 8:23
• @Zelazny I made a mistake in my comment. The equivalence class of $\mathbb{N}$ will contain every "infinite" subset of $\mathbb{N}$. For example, $\mathbb{N} \equiv \{10, 20, 30, \ldots\} \equiv \{1,23,56,987,\ldots\}$. So the size of this equivalence class will not be $1$ as I indicated in my previous comment. – Anurag A Jan 13 '17 at 8:08