# Prove Relationship between Equivalence Relations and Equivalence Classes

1. Oct 13, 2011

### Ceci020

I'm not sure if I did these 2 questions correctly, so would someone please check my work for any missing ideas or errors?

Question 1:
1. The problem statement, all variables and given/known data
Prove:
For every x belongs to X, TR∩S(x) = TR(x) ∩ TS(x)

2. Relevant equations

3. The attempt at a solution
TR(x) = {x belongs to X such that <x,y> belongs to R}
TS(x) = {x belongs to X such that <x,y> belongs to S}
TR∩S(x) = {x belongs to X such that <x,y> belongs to R∩S}

<x,y> belongs to R∩S if <x,y> belongs to R and also belongs to S, which satisfy the definition above for TR(x) and TS(x)

Question 2:

1. The problem statement, all variables and given/known data
R and S are equivalence relations over X
Prove R ∩ S is also an equivalence relation over X

2. Relevant equations

3. The attempt at a solution
Since R and S are equivalence relations over X, then for x in X, R and S satisfy properties:
Reflexive:
<x,x> belongs to R
<x,x> belongs to S
Symmetry:
<x,y> and <y,x> belong to R
<x,y> and <y,x> belong to S
Transitivity:
<x,y> belongs to R, <y,z> belongs to R; then <x,z> belongs to R
<x,y> belongs to S, <y,z> belongs to R, then <x,z> belongs to S

If R∩S is equivalence relation, then it must satisfy:
1/ <x,x> belongs to R∩S, meaning <x,x> belongs to R and also belongs to S
2/ <x,y> and <y,x> belong to R∩S, meaning <x,y> belongs to R and also belongs to S.
3/ <x,y> belongs to R∩S and <y,z> belongs to R∩S, then <x,z> belongs to R∩S, meaning <x,z> belongs to R and also belongs to S
All of these are satisfied by hypotheses.
So R∩S is equivalence relation over X.

2. Oct 13, 2011

### HallsofIvy

Staff Emeritus
First, be carful with your wording:
should be "For all x in X, <x,x> belongs to R and <x,x> belongs to S"
And you really should say "therefore <x, x> belongs to $R\cap S$".

would normally be interpreted as "for all x and y in X, <x, y> and <y, x> belong to R", etc. but that is not what you want to say. IF <x, y> is in R, then <y, x> if in R.

and the same for "transitive": IF <x, y> is in R AND <y, z> is in R, then <x, z> is in R.

And you cannot just say "all of these are satisfied by hypotheses". You must show exactly why each of reflexive, symmetric, and transitive is satisfied for $R\cap S$.