Question

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

Answer 1

Hint: 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.