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The question is, "Show that the relation R = ∅ on the empty set S = ∅ is reflexive, symmetric, and transitive."

I was told by my teacher that you could simply say it can't be shown that each property isn't true; and that would show that the relation had those three properties. To me, this answer isn't very satisfying. Could someone, perhaps, elaborate on this idea more?

Thank you!

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To show reflexivity, note that for every $x\in\varnothing$, we have $xRx$.

To show symmetry, note that for every $x,y\in\varnothing$, we have $xRy$ implies $yRx$.

To show transitivity, note that for every $x,y,z\in\varnothing$, we have $xRy$ and $yRz$ implies $xRz$.

These are vacuously true because the empty set contains no elements.

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    $\begingroup$ To elaborate on the "vacuously true" statement: To show that a relation is not reflexive on a set $X$, you need to show that there is some $x\in X$ such that $x\not\mathrel{R} x$. But this is impossible if $X=\emptyset$, since there are no $x\in X$, period. Similarly for the other two properties. $\endgroup$ – Andrés E. Caicedo Nov 3 '12 at 17:30
  • $\begingroup$ Hmm...are you assuming $x\in\varnothing$ is true? Are we allowed to assume that x is an element of the null set? $\endgroup$ – Mack Nov 3 '12 at 17:32
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    $\begingroup$ I think this is what the instructor meant by "it cannot be shown that the property is not true", since showing that it fails requires something impossible: Finding elements of the empty set (with additional properties). $\endgroup$ – Andrés E. Caicedo Nov 3 '12 at 17:36
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    $\begingroup$ @EMACK: Think about it this way: how would you prove that $R$ is not reflexive? You’d have to find an $x\in S$ such that $\langle x,x\notin R$. But you can’t find an $x\in S$ in the first place, so you certainly can’t find one that has some particular property of interest. It’s vacuously true that $xRx$ for all $x\in\varnothing$ simply because there is no possible counterexample. $\endgroup$ – Brian M. Scott Nov 3 '12 at 17:46
  • $\begingroup$ I see, now. Thank you, everyone, for your help! $\endgroup$ – Mack Nov 3 '12 at 18:42

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