Need help to understand equivalence class

Question

This is in my note Let S={1,2,3,4} Let R be the relation on P(s) defined by xRy <=>|x|=|y|

how many equivalence classes are there ?

5

[∅]={∅}

[{2}]={{1},{2},{3},{4}}

[{2,3}]={{1,2},..........

[a]=the set of all element of A that are related to A

[{!}]=the set of all elements of P(s) that are related to {1}.

how do you find the number on the left [{2,3}],[∅]??? If you can explain equivalence class to me in other way. ty!

Answer 1

I think, the particular classes written $[\{2\}]$ and $[\{2,3\}]$ are only examples. The main point is, I guess, clear, an $n$ element subset of $S$ is related exactly to the $n$ element subsets by $R$.

In general, if $M$ is a set and $R$ is an equivalence relation on $M$, then the quotient set $M/R$ which formally consists of the equivalence classes, can also be viewed as the realization of having all the elements of $M$ with the original equality replaced by the relation $R$. So that, each element $m\in M$ is (determines) an element in $M/R$, and $m=m'$ holds in $M/R$ iff $\ m\,R\,m'$ in $M$. Formally, to distinguish, we should rather write it using brackets, like $[m]=[m'] \iff m\,R\,m'$.

Another important perspective is equivalence relations via surjections. Every equivalence relation on $M$ can be defined by a (surjective) function $f:M\to K$ onto some set $K$:

Let $x\,E_f\,y$ iff $\ f(x)=f(y)$.

Then, this same $f$ determines a bijection between $M/E_f$ and $K$, in other words, in this case the quotient set can be identified with the range of $f$ ($K$), via the mapping $f$, so that each value of $K$ will straightly correspond to one equivalence class.

In this particular example, for $A\in P(S)$, we can set $f(A):=|A|$, then $R=E_f$, and now it takes values from the range $\{0,1,2,3,4\}$.