The given equivalence relation is defined as $r \subseteq P(\mathbb{N})^2$:
$P \ r \ Q$ if and only if $P = Q = \emptyset$
or
$P, Q \neq \emptyset$ and $\min P = \min Q$.
What is the cardinality of the quotient set $P(\mathbb{N})/_r$? What are the cardinalities of individual abstraction classes?
Don't really understand what is the cardinality of the quotient set $P(\mathbb{N})/_r$? What does $P(\mathbb{N})/_r$ mean? For second question I should find and write some individual abstraction classes and then, depending on these classes, i should find their cardinalities?
Any help would be much appreciated.
The quotient set $P(\mathbb{N})/_r$ is just the set of all the equivalence classes modulo $r$. What are they? Each class contains subsets of $\mathbb{N}$ that have the same minimal element. Note that there are as many such equivalence classes as there are possible minimal elements, i.e. natural numbers. So the cardinality of $P(\mathbb{N})/_r$ is just $\aleph_{0}$, i.e. the cardinality of $\mathbb{N}$. The bijection to show that $P(\mathbb{N})/_r$ is countable is straightforward.
A formal argument showing the cardinality of each equivalence class will require more work. But here is a start: the class containing the empty set is a singleton, and note that most (if not all) other equivalence classes will be equinumerous with $P(\mathbb{N})$ because they're essentially the powersets of sets equinumerous with $\mathbb{N}$, e.g. $\{ 1,2,3,... \}$, $\{ 2,3,4,... \}$, $\{ n,n+1, n+2,... \}$. For instance, take the equivalence class $[n]$. It is just the powerset of $\{ n+1,n+2,... \}$ such that each element $S$ of $P(\{ n+1,n+2,... \})$ is appended with $n$, i.e. $S \cup \{ n \}$.
Define the equivalence classes as $[n]:= \{S \in P(\mathbb{N}) : \mathrm{min}(S) = n \}$. And there's also the class $[\emptyset] = \{ \emptyset \}$. Now, $P(\mathbb{N})/_r = \{[n]:n \in \mathbb{N} \} \cup \{ [\emptyset] \} = \{ [\emptyset],[0],[1],[2],... \}$. Define the bijection $f:P(\mathbb{N})/_r \longrightarrow \mathbb{N} $ as follows: $f([\emptyset])=0$ and $f([n])=n+1$. All you need to show is that it is both injective and surjective, which are elementary arguments.
As for the cardinalities of each $[n]$ you could first recharacterize them in a way as to make their cardinalities clear, that is: $[0]= \{S \in P(\mathbb{N}):0 \in S \}$ and $[n]= \{ S \in P(\mathbb{N} \backslash \{ 0,1,...,n-1 \}):n \in S \}$, so therefore each $[n]$ is equinumerous with $P(\mathbb{N})$, since it is the powerset of a countably infinite set, i.e. $\mathbb{N}$ with a truncated initial (finite) segment.
It would be even more clear that each $[n]$ has cardinality $P(\mathbb{N})$, i.e. is uncountable, if we characterized each $[n]$ in a manner where we explicitly add, rather than subtract from some set clearly equinumerous with $P(\mathbb{N})$. For instance we could use the following characterization instead: $[n]= \{S \cup \{n \}: S \in P(\mathbb{N} \backslash \{ 0,1,...,n \})$.