Consider $A=^*B\iff A\triangle B$ is finite ($\triangle$ is the symmetric difference) what type of relation is $=^*$
First i start by checking if it is reflexive, anti-symmetric and transitive.
so for reflexive i have put let A be a set then $A=^*A \iff A\triangle A$ is finite so $A\triangle A=(A\setminus A)\cup (A\setminus A)=\emptyset$ which is finite (since $(A\setminus A)=\emptyset$) thus $A=^*A$ so the reflexive property holds?
Now for anti-symmetric: if $A=^*B$ and $B=^*A$ then $A=B\iff A\triangle B \land B\triangle A$ are finite, now i've gone on to show that $A\triangle B= B\triangle A$ but how do i show that they are finite? (or is it sufficent to show that they are equal and state that $A,B$ non-empty but if they was empty then surely they would just both be equal to $\emptyset$ .)
Now for the transitive property : If $A=^*B$ and $B=^*C$ then $A=^*C$ would i use the result that $A\triangle B$=$B\triangle A$ ? i'm not too sure what to do here.
if it was to satisfy these three properties then it would be a partial ordering and if it satisfies the Trichotomy law then it would be a total ordering.
Any help would be extremely appreciated.