I have the assignment:
Provide a direct proof of this statement: Let A and B be two arbitrary sets such that B ⊆ A, and let R be an equivalence relation on A. Consider the relation S = {(a, b) ∈ R | a, b ∈ B} on the set B. Then S is an equivalence relation on B (that is, reflexive, symmetric and transitive).
Obviously i have to show that S is reflexive, symmetric and transitive. But how would this be done?
Any tips would be greatly appreciated
We have a relation $R\subset A\times A$ of pairs of elements of $A$. $S$ however, is exactly those pairs in $R$ that have both elements in $B\subset A$.
Since $R$ is reflexive, for any $a\in A$ we have $(a,a)\in R$. In particular for any $b\in B\subset A$ we have $(b,b)\in R$. But this is a pair with both elements in $B$, so $(b,b)\in S$ for any $b\in B$. Therefore $S$ is reflexive.
Since $R$ is symmetric, if $(a,a')\in R$ for some $a,a'\in A$, then also $(a',a)\in R$. Now, if $(b,b')\in S\subset R$, then also $(b',b)\in R$ as $R$ is symmetric, but again $(b',b)$ is a pair with both elements in $B$, so $(b',b)\in S$, meaning that $S$ is symmetric.
For transitive, let $b,b',b''\in B\subset A$ and $(b,b')\in S$ and $(b',b'')\in S$. Then $(b,b')$ and $(b',b'')$ are both elements of $R$, since $S\subset R$. Therefore $(b,b'')\in R$, because $R$ is transitive. But this is a pair with both elements in $B$, so $(b,b'')\in S$. This shows that $S$ is transitive.
