Doubt

Question

A $\phi$ (empty) relation on any set $A$ is not reflexive because for every $ a \in A$, $(a, a) \notin \phi$, but $\phi$ is a symmetric as well as transitive relation on $A$, how is that possible $?$

Comments

Since their is no (a,b) pair present and no  (a,b) , (b,c) is present.. So relation is symitric and transitive.. Min  Cardinality of symetrric and transitive relation is 0 . but for reflexive it is n.. (Assume n no. Is present.)

Answer 1

If R is relation defined on set X, then,  

• For reflective property on a non-empty set X, we need the following condition.

          $$\forall x \in X , (x,x) \in R$$

• $\phi$ does not satisfy this. $\Rightarrow$ NOT Reflexive.

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• For symmetric property on a set X, we need the following implication hold.

          $$\forall x \forall y \Big((x,y)\in R \to (y,x)\in R\Big)$$

• for $\phi$ relation $(x,y)\in R$ is false $\Rightarrow$ Above implication is trivially true.

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•  For transitive property on a set X, we need the following implication hold.

          $$\forall x\forall y\forall z \Big((x,y)\in R\wedge (y,z)\in R \to (x,z)\in R\Big)$$

• for $\phi$ relation $(x,y)\in R\wedge (y,z)\in R$ is false.$\Rightarrow$ Above implication is trivially true.

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NOTE: for empty set X, empty relation $\phi$ is also reflexive.

Answer 2

For a relation to be reflexive on a set A, all pairs of the form (a,a) should be present in the relation. Let's consider $ A = \big \{1,2,3 \big\}$ here. Then for a relation to be reflexive all pairs $(1,1) , (2,2)$ and $(3,3)$ must be present in the relation.

So, $R_{1} = \big \{(1,1),(2,2),(3,3) \big\}$, $R_{2} = \big \{ (1,1),(2,2),(3,3),(1,3),(2,3) \big \}$ are reflexive while $R_{3} = \big \{(1,1),(2,2) \big\}$ and $R_{4} = \phi $ are not.

And for a relation to be symmetric there is a condition that if $a^Rb$, then $b^Ra$ $\forall a,b \in A$, Now, In case of relation $\phi$, the first condition i.e., $a^Rb$ is not met, So we don't proceed further to check whether $b^Ra$ (or) not.
And Similarly In case of transitive relation if  $a^Rb$ and $b^Rc$ then only we check whether $a^Rc$ (or) not. If the first two conditions are not met, there is no need to check whether $a^Rc$ (or) not.

Thus, relation $\phi $ is Symmetric, Transitive but not reflexive.