Is an empty set reflexive? Symmetric? Transitive?

Question

Suppose

$$A\neq\emptyset$$

Since, $$\emptyset\subseteq A\times A$$

the set $$R=\emptyset$$ is a relation on A.

Is $R$ reflexive? symmetric? transitive?

I remember hearing something can be "vacuously" true. So the empty set would be reflexive, symmetric and transitive because it doesn't meet the definition?

So there is no $(x,x)$ that can exist in $R$ therefore vacuously reflexive.

There is no $(x,y)$ that can exist in $R$ therefore vacuously symmetric.

There is no $(x,y)$ that can exist in $R$ therefore vacuously transitive.

Is my reasoning correct here?

Look at the specific definitions of reflexivity, transitivity and symmetry for a relation. Do they apply to the empty set? — Steven Stadnicki, Mar 24 '15 at 02:44
Doesn't reflexive mean "For every $x$, if $x$ is in $A$, then $(x,x)$ is in $R$"? This is clearly false: If $A\neq{}$, then it has elements $x$, and clearly $(x,x)$ can't be in ${}$. — Akiva Weinberger, Mar 24 '15 at 02:49
As for symmetry: "For every $x$ and $y$, if $(x,y)$ is in $R$, then $(y,x)$ is in $R$." Well, the statement "if $(x,y)$ is in $R$" is false for every possible $x$ and $y$. False implies true in math, so... Similar stuff for the last one. (P.S. I don't know why the downvote; this is a legitimate question.) — Akiva Weinberger, Mar 24 '15 at 02:52

score 6 · Accepted Answer · edited Jan 11 '22 at 17:53

I think its transitive automatically because the relation only has the empty set but I'm not sure.

The term is "vacuously".

A relation is transitive if $\forall x\forall y\forall z \Big((x,y)\in R\wedge (y,z)\in R \to (x,z)\in R\Big)$.

This is vacuously true because you cannot find any counterexamples, since the relation is empty. (The implication is never falsifiable)

So there is no $(x,x)$ that can exist in $R$ therefore vacuously reflexive

No, the set $A$ is not empty, so $\forall x( x\in A\to (x,x)\in R)$ is not a vacuous truth; it is in fact fallacious.

However, the definition for irreflexive is $\forall x( x\in A\to (x,x)\notin R)$, so that is true, although not vacuously so.

There is no $(x,y)$ that can exist in $R$ therefore vacuously symmetric

Yes, symmetry requires $\forall x \forall y \Big((x,y)\in R \to (y,x)\in R\Big)$.

Now, what about antisymmetry, and asymmetry?

There is no $(x,y)$ that can exist in $R$ therefore vacuously transitive

Done

Thanks I've edited my post since I remembered the term "vacuously" true — Hare Krishan, Mar 24 '15 at 02:51

Is an empty set reflexive? Symmetric? Transitive?

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