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Can someone help me understand the properties of the relation $R = \emptyset$ ? It looks to me like it's not reflexive, since there is no element related to any element, so the elements are not related to themselves ! Right ? So, does it mean it's irreflexive ?

About symmetry, I guess I understand it is symmetric, since there is not element related to another..

Same thing makes it's asymmetric and antisymmetric ? Or not necessarily ?

Finally, I don't seem to get why it'd be transitive ! If there is no element !

I hope someone has a clear way of understanding this topic. Thanks !

Sorry, I forgot to add that it's a relation on $N^2$ ... therefore, we can say it's reflexive, symmetric, antisymmetric and transitive. It's not irreflexive and it's not asymmetric ? Right ? thanks to you all !!

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    $\begingroup$ When seen as a relation on the empty set, it is all of the above, But when seen as a relation on a non-empty set, at least one of them fails... $\endgroup$ – Thomas Andrews Feb 26 '14 at 0:18
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Whether the empty relation is reflexive or not depends on the set on which you are defining this relation -- you can define the empty relation on any set $X$.

The statement "$R$ is reflexive" says: for each $x\in X$, we have $(x,x)\in R$. This is vacuously true if $X=\emptyset$, and it is false if $X$ is nonempty.

The statement "$R$ is symmetric" says: if $(x,y)\in R$ then $(y,x)\in R$. This is vacuously true, since $(x,y)\notin R$ for all $x,y\in X$.

The statement "$R$ is transitive" says: if $(x,y)\in R$ and $(y,z)\in R$ then $(x,z)\in R$. Similarly to the above, this is vacuously true.

To summarize, $R$ is an equivalence relation if and only if it is defined on the empty set. It fails to be reflexive if it is defined on a nonempty set.

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  • $\begingroup$ Quite correct. I've edited my post. $\endgroup$ – bradhd Feb 26 '14 at 2:19
  • $\begingroup$ Is the empty relation $R$ on $\emptyset$ alsy antisymmetric (making it a total order and an equivalence relation)? $\endgroup$ – YoTengoUnLCD Aug 4 '15 at 0:58
  • $\begingroup$ Confused due to first line. What is the answer (true/false) if asked "Empty set $\phi$ is an equivalence relation"? That is when we are not told with which set this empty set is associated with. Do we always have to define empty set in relation with some other set? $\endgroup$ – anir123 Feb 4 '17 at 11:46
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The basic fact here is that an implication "if $P$ then $Q$" is false in one case only: that where $P$ is true and $Q$ is false. For example, to show that the statement $$\hbox{"if $x$ is a bird then $x$ can fly"}$$ is not always true, you have to give an example of something which is a bird but cannot fly.

The statement that $R$ is transitive is that for all $x,y,z$, $$\hbox{if $x\,R\,y$ and $y\,R\,z$ then $x\,R\,z$.}$$ Now if $R$ is the empty relation, then the first part $$\hbox{$x\,R\,y$ and $y\,R\,z$}$$ cannot be true, so the statement as a whole cannot be false. Therefore $R$ is transitive. We often say that $R$ is vacuously transitive.

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A relation, $R$, is symmetric when for all $x,y$, if $(x,y)\in R$ then $(y,x)\in R$. But here, since $R$ is empty, it has no elements $(x,y)$, so the hypothesis is empty. But the conclusion of an implication is true even with an empty hypothesis. So, $R$ is symmetric.

Same goes for transitivity.

But it is not reflexive, because $(x,x)\notin R, \forall x$, since $R$ is empty. Only in the case that $X=\emptyset$, then again because of empty hypothesis, we have that $R$ is reflexive.

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  • $\begingroup$ @Ink Yes, you're right. I'll add this. $\endgroup$ – frabala Feb 26 '14 at 10:25

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