Question

Prove that intersection of equivalence relations on a set is also an equivalence relation.

Answer 1

An equivalence relation is a binary relation that is reflexive, symmetric, and transitive. For the objects where
a=a (reflexive property)
if a=b and b=a(symmetric property)
if a=b and b=c, then a=c (transitive property)
In this question, to prove that the intersection of equivalence relations on a set is also an equivalence relation proves all that the set satisfies all the properties of equivalence.

Complete step-by-step solution
Let us assume P and Q are two equivalence sets whose intersection is to be proved of set S
So set S implies $P \cap Q$ equivalence, where
$x \in S \Rightarrow \left( {x,x \in P} \right)$and$x \in S \Rightarrow \left( {x,x \in Q} \right)$
$\left( {x,x} \right) \in P \cap Q$
$\therefore P \cap Q$is Reflexive
Now to check symmetric
$\left( {x,y} \right) \in P \cap Q$
Hence
$\left( {x,y} \right) \in P$and $\left( {x,y} \right) \in Q$
are symmetrical
$\left( {y,x} \right) \in P$and $\left( {y,x} \right) \in Q$
$\left( {y,x} \right) \in P \cap Q$
Hence $P \cap Q$is symmetric.
Now for transitive property
$\left( {x,y} \right) \in P \cap Q$and $\left( {y,z} \right) \in P \cap Q$
$\left( {x,y} \right) \in P$and\left( {y,z} \right) \in P$$$$ \Rightarrow \left( {x,z} \right) \in P
$\left( {x,y} \right) \in Q$and\left( {y,z} \right) \in Q$$$$ \Rightarrow \left( {x,z} \right) \in Q
Therefore P and Q are transitive
$P \cap Q$is transitive
Hence all the properties of equivalence are satisfied, therefore $P \cap Q$is an equivalence relation.

Note: Two elements from an equivalence relation are called equivalent. The set of one element has only one equivalence relation with one equivalence class.