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Is a relation defined by an emptyset symmetric and transitive?

It is obviously not reflexive, but is it symmetric and transitive? My thinking is it is, as there is no example of it not being either. For example, there is no example of $aRb \not \Leftrightarrow bRa$ or $(aRb) \wedge (bRc) \not \Rightarrow aRc$ in the empty set. Sorry if this is a trivial question, I am still wrapping my head around relations as a topic

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    $\begingroup$ Welcome to Mathematics Stack Exchange. It is vacuously $\endgroup$ – J. W. Tanner Jul 26 '20 at 2:54
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All three conditions can be cast as conditional statements:

  • Reflexive:$\;$If $a\in X$, then $aRa$.$\\[4pt]$
  • Symmetric:$\;$If $a,b\in X$ and $aRb$, then $bRa$.$\\[4pt]$
  • Transitive:$\;$If $a,b,c\in X$ and $aRb$ and $bRc$, then $aRc$.

For any set $X$, if $R$ is the empty relation on $X$, the hypothesis is false for the symmetric and transitive conditionals, hence those statements are true.

For the reflexive conditional, if $X\ne\large\varnothing$ and $R$ is the empty relation on $X$, then by choosing $a\in X$, the hypothesis is true but the conclusion is false, hence the statement is false.

Thus if $X\ne\large\varnothing$ and $R$ is the empty relation on $X$, $R$ is symmetric and transitive but not reflexive.

However if $X=\large\varnothing$ and $R$ is the empty relation on $X$, the hypothesis for each of the three conditionals is false, hence those statements are true.

Thus the empty relation on the empty set qualifies as an equivalence relation.

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