R is a relation over P(S): R = {(C, D) ∈ P (S)^{2} : C ∩ E = D ∩ E} is R an Equivalence relation?

Question

given a group S,E ⊆ S . we shall define a relation R, over P(S):

R = {(C, D) ∈ P (S)^${2}$ : C ∩ E = D ∩ E}

I need to prove that this relation is a Equivalence relation, but I'm confused about the way this relation is defined.

help will be appreciated very much.

Answer 1

For any function $f:X\to Y$ the relation $x\sim y\iff f(x)=f(y)$ is an equivalence relation on $X$. This is easy to prove. E.g. the fact that $\forall x\in X [f(x)=f(x)]$ already tells us immediately that the relation is reflexive. Also symmetry and transitivity are easy.

Prescribe function $f:\wp(S)\to\wp(S)$ by $C\mapsto C\cap E$ and apply.

Answer 2

An equivalence relation on a set $X$ is a subset $R$ of $X\times X$ satisfying three axioms:

• reflexivity: for any $x\in X$, $(x,x)\in R$

• symmetry: for any $x,y\in X$, if $(x,y)\in R$ then $(y,x)\in R$

• transitivity: for any $x,y,z\in X$, if $(x,y), (y,z)\in R$ , then $(x,z)\in R$.

For any $x,y\in X$, in general, we write $x R y$ or $x\sim_R y$ instead of $(x,y)\in R$. You may now relate the previous definition to the definition I suppose you know.

In your case, this translates as follows: for any two subsets $C,D$ of $S$, we have $C \sim_R D$ if and only if $C\cap E=D\cap E$. Nothing more, nothing less.

I let you check that you get indeed an equivalence relation.