Equivalence relation and restriction

Question

This is a HW question

Suppose $B \subseteq A$ and $R_a$ is an equivalence relation on A. Let $R_b$ the restriction of $R_a$ to B; that is, $R_b = \{(a,b) \in R_a : a,b \in B\} $ Is $R_b$ an equivalence relation on B.

I know that for am equivalence relation I need the the relation to be reflexive, antisymmetric and transitive.

To me this seems too simple and usually when i think of a HW question that way I am wrong. My reasoning is that if $(a,a) \in A$ then $(a,a) \in B $ because $B \subseteq A$ therefore $R_b$ is reflexive. Then similarly the anti symmetric and transitive properties can be shows.

Answer 1

Your reasoning is not quite right. In order to show that $R_b$ is an equivalence relation on $B,$ you must show the following:

• For every $x$ in $B,$ $(x,x)$ is in $R_b$.
• If $x,y$ are in $B$ and $(x,y)$ is in $R_b,$ then $(y,x)$ is in $R_b$.
• If $x,y,z$ are in $B$ and $(x,y),(y,z)$ are in $R_b,$ then $(x,z)$ is in $R_b$.

It is fairly straightforward to show each of these using the definition of $R_b,$ since $R_a$ is an equivalence relation on $A$ and $B\subseteq A$. Let me prove the second property as an example, and leave the other two to you.

Suppose $x,y$ are in $B$ and $(x,y)$ is in $R_b$. Since $B\subseteq A$ and $x,y$ are in $B,$ then $x,y$ are in $A$. Since $(x,y)$ is in $R_b,$ then by definition of $R_b,$ we have $(x,y)$ is in $R_a.$ Since $x,y$ are in $A$ with $(x,y)$ in $R_a,$ and $R_a$ is an equivalence relation on $A,$ then $(y,x)$ is in $R_a$ by symmetry of $R_a$ on $A$. Since $y,x$ are in $B$ and $(y,x)$ is in $R_a,$ then by definition of $R_b,$ we have that $(y,x)$ is in $R_b,$ as desired.

Note that no mention was made of ordered pairs in $A$ or $B$. (1) There's no reason to suspect that there are any such things. (2) Even if there are ordered pairs in $A$ or $B$, that is not connected to ordered pairs in $R_a$ or $R_b,$ so has nothing to do with the problem at hand. (3) Even if there is some ordered pair in $A,$ we can't conclude that that ordered pair is in $B$ ($B$ is the subset here).