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!


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.

  • 3
    $\begingroup$ 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
  • 4
    $\begingroup$ @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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