Show that the restriction of an equivalence relation is an equivalence relation. Let $C$ be a relation on a set $A$. If $A_{0} \subset A$, define the restriction of $C$ to $A_{0}$ to be the relation $C \cap (A_{0} \times A_{0})$. Show that the restriction of an equivalence relation is an equivalence relation.
$\textit{Proof Attempt}$
Let's assume $C$ is an equivalence relation then it satisfies the 3 properties (reflexivity, Commutativity and transitivity).
Now $(C \cap (A_{0} \times A_{0})) \subset C$ hence $(C \cap (A_{0} \times A_{0}))$ also satisfies the 3 properties (reflexivity, Commutativity and transitivity).
Since $C$ and $A_{0}$ are arbitrary we conclude the restriction of an equivalence relation is an equivalence relation.
 A: Unfortunately, none of the properties of reflexivity, symmetry (the usual term for what you are calling "commutativity"), nor transitivity is inherited by subsets. 
Moreover, note that while symmetry and transitivity are not contextual (they depend only on your set of pairs, that is, on your relation), reflexivity is contextual: you need to say what your underlying set is in order to discuss reflexivity: the exact same set of pairs may be reflexive when considered as a relation on a set $A$, but not when considered as a relation on a different set $B$, even if $C\subseteq (A\times A)\cap(B\times B)$. 
That is, it is false that if $D\subseteq C$ and $C$ is reflexive (on something), then $D$ is reflexive (on something else). It is false that if $D\subseteq C$ and $C$ is symmetric, then $D$ is symmetric. And it is false that if $D\subseteq C$ and $C$ is transitive, then $D$ is transitive.
For instance, take $A=\{1,2,3\}$, $C=\{(1,1), (1,2), (2,1), (2,2), (3,3)\}$, $D=\{(1,1), (1,2),\}$. Then $C$ is reflexive, symmetric, and transitive, $D\subseteq C$, but $D$ is not reflexive (not on $A$ and not on $\{1,2\} $) and not symmetric. You can probably come up with examples where it is not transitive either.
Of course, I didn't construct $D$ by restriction; but the point is that nowhere did you use the fact that you were constructing your relation by restriction. You just asserted that $C\cap(A_0\times A_0)$ would have the desired properties by virtue of being contained in a relation that did, and that argument is invalid.
What you need to answer:


*

*Reflexivity on $A_0$. Let $a\in A_0$; why must $(a,a)$ be in $C\cap (A_0\times A_0)$ (Hint: Must it be in each of the two sets you are intersecting?)

*Symmetry. If $(a,b)\in C\cap(A_0\times A_0)$, why must $(b,a)$ also be in $C\cap (A_0\times A_0)$?

*Transitivity. If $(a,b)\in C\cap (A_0\times A_0)$, and $(b,c)\in C\cap(A_0\times A_0)$, why must $(a,c)$ be in $C\cap(A_0\times A_0)$?
