# Followup on Proof: Equivalence Classes

The original question at hand was:

Suppose α and β are equivalence relations on the set S. Suppose further that the relation γ is defined as follows: For x,y∈S, xγy means xαy and xβy. Prove that γ is also an equivalence relation on S.

I'm not quite sure where I should start on this, but I think it has something to do with both α and β being equivalence relations, and then proving that γ is also an equivalence relation through the use of reflexive, symmetric, and transitive properties. Any ideas?

The followup to this question is: Suppose, in the preceding problem, S = {1, 2, 3, 4, 5, 6}, the equivalence classes of α are {1, 2, 3} and {4, 5, 6} and the equivalence classes of β are {1, 4}, {2, 5}, and {3, 6}. What are the equivalence classes of γ?

Any ideas on this one? I have no idea how to approach this.

• Just go through the axioms one by one. For example we have $x\gamma x$ if and only if $x \alpha x$ and $x\beta x$. But we know... – leibnewtz Nov 17 '18 at 18:53
• Could you give an example of this? – NKP Nov 17 '18 at 19:30

For the first question, check the axioms one by one. For example (i) reflexive: Take any $$x\in S$$. $$x\alpha x$$ and $$x\beta x$$, thus... (continue)
For the second part, start with one element in $$S$$, e.g. with $$1$$ and then add all elements which are equivelant to $$1$$. If you have all of them, continue and do the same thing again. This is a valid strategy, since $$S$$ is finite. Make sure that you understand why an equivalence relation partitions its set into classes.