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.

  • 2
    $\begingroup$ It sounds like a badly phrased question or trick question. In case you're translating from another language, are you sure it doesn't ask what type of relation $=^*$ is? In that case the answer would be that it's an equivalence relation -- but one wouldn't usually call that a "type of ordering". $\endgroup$ – Henning Makholm Oct 7 '17 at 23:42
  • $\begingroup$ You're correct, it says what kind of relation sorry, i couldn't read my supervisors hand writing, luckily since you say it is an equivalence relation then instead of trying to show anti-symmetry holds i would just show symmetry holds? $\endgroup$ – user395952 Oct 7 '17 at 23:45
  • $\begingroup$ In fact you should be able to prove that it's not antisymmetric fairly easily by showing a concrete counterexample. $\endgroup$ – Henning Makholm Oct 7 '17 at 23:46
  • $\begingroup$ i've only been able to show $A\triangle B$=$B\triangle A$ i cant seem to think of an example that is not finite. could you enlighten me? $\endgroup$ – user395952 Oct 7 '17 at 23:48
  • $\begingroup$ A counterexample to "antisymmetric" would be sets $A$ and $B$ such that $A=^*B$ and $B=^*A$ yet $A\ne B$. Why do you think you would need to find an infinite symmetric difference for that? $\endgroup$ – Henning Makholm Oct 7 '17 at 23:49

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