Is a relation, R, an Equivalence Relation of a Power Set? Where $A = \{1,2,3,4,5,6\}$ and $S = P(A)$ is the power set, for $a,b \in S$ define a relation $R: (a,b) \in R$ where $a$ and $b$ have the same number of elements.
Is $R$ an equivalence relation on $S$ and if so how many equivalence classes are there?
I've defined my $R$ as being $\{(\{1,2\},\{3,4\}),(\{1,2\},\{5,6\}),(\{1,2\},\{1,2\}),(\{3,4\},\{1,2\}),(\{3,4\},\{3,4\}),(\{3,4\},\{5,6\}),(\{5,6\},\{1,2\}),(\{5,6\},\{3,4\}),(\{5,6\},\{5,6\})\}$
because of the part of the question mentioning $a$ and $b$ have the same number of elements. Was that wrong to do?
 A: In general if $f:X\to Y$ is a function then the relation $R$ on $X$ defined by: $$uRv\iff f(u)=f(v)$$ is an equivalence relation on $X$. 
Equality $f(u)=f(u)$ guarantees reflexivity. 
Implication $f(u)=f(v)\implies f(v)=f(u)$ guarantees symmetry.
Implication $f(u)=f(v)\wedge f(v)=f(w)\implies f(u)=f(w)$ guarantees transitivity.
It is always possible (and handsome) to let $f$ be surjective by restricting its codomain. Then equivalence classes are the fibres $f^{-1}(\{y\}):=\{x\in X\mid f(x)=y\}$ for $y\in Y$ and consequently the cardinality of $Y$ equals the number of equivalence classes.
You can apply this here on function $f:S\to\{0,1,2,3,4,5,6\}$ prescribed by:$$s\mapsto\text{number of elements of }s$$
This function is surjective and the cardinality of $\{0,1,2,3,4,5,6\}$ is $7$. So there are $7$ equivalence classes.
A: Yes that was wrong to do since you missed a lot of sets for example
$$
(\{1,2,3\},\{2,3,4\})\in R
$$
$R$ is indeed an equivalence relation (appeal to the definition) and any two members of an equivalence class have same number elements. There are $7$ possible cardinalities including $0$ and hence there are $7$ equivalence classes.
