Transitivity and symmetry does not imply reflexivity counterexample?

Question

I am not understanding the following counterexample (found in a solutions manual) for a transitive and symmetric relation that is not reflexive.

Consider set $A = \{1,2,3\}$. Then $R = \{(1, 3), (3, 1), (1, 1), (3, 3)\}$ is symmetric and transitive, but not reflexive. The definition of a symmetric relation is for all elements x and y in set A, if $xRy$ then $yRx$. Clearly 2 is an element of A, but it is not symmetric according to the definition. And doesn't the existence of $(1, 1)$ and $(3, 3)$ point to reflexivity of R?

Answer 1

Don't just bold the "for all elements". Bold the if.

Definition of "symmetric"

"For all elements $x$ and $y$ in the set:" This means $(x,y)$ can be $(1,1)(1,2)(1,3)(2,1)(2,2)(2,3)(3,1)(3,2)(3,3)$.

of those

"If $xRy$ ..." This means $(x,y)$ can be $(1,1),(3,1),(1,3),(3,3)$ neither $x$ nor $y$ can be $2$ because we don't have any $2Ry$ or $xR2$.

"then $yRx$". $1R1$ and $1R1$. check. $3R1$ and $1R3$. check. $1R3$ and $3R1$. check. $3R3$ and $3R3$. check.

So R is symmetric. It doesn't matter that $2 \not R y$ because $2 \not R y$. We only need to check IF $2R y$ then $yR2$ is always true. But as $2Ry$ NEVER occurs, this is vacuously true.

Now, definition of reflexive.

For all $x$ in the set, we will have $xRx$. Let's check: $1R1$, check. $2R2$, FAIL. $3R3$, check. It fails and is not reflexive.

So, Transitive and symmetric does not imply reflexive.

But Transitive and symmetric AND $\forall x, \exists y| xRy$ does imply reflexive.

(As for all $x$ there is a $y$ so that $xRy$. So by symmetry $yRx$ so by trasitivity $xRy$ and $yRx$ so $xRx$.)

Also if R is transitive and symmetric but NOT reflexive there MUST exists some $x$ where $x$ is not related to anything. In this counter example, that element is $2$. $2$ is not related to anything.

Answer 2

The symmetry condition is a universally quantified implication. It tells us that for all $x, y \in A$ we must have that if $x R y$ then $y R x$. But there are no $z \in \{1,2,3\]$ such that $zR2$ or $2Rz$, so this condition is vacuously satisfied.

The reflexivity condition is also universally quantified but does not involve implication. It tells us that for all $x \in A$ we must have that $xRx$. But since there is no $z \in \{1,2,3\]$ such that $zR2$ or $2Rz$, this does not hold.

Answer 3

Consider the null set as your relation.