Which defination is correct? (symmetric relation)

Question

My textbook(2nd page, at very bottom) states the definition of symmetric relation as follows:

A relation $R$ in a set $A$ is called symmetric, if $\left(a_{1}, a_{2}\right) \in \mathrm{R}$ implies that $\left(a_{2}, a_{1}\right) \in \mathrm{R},$ for all $a_{1}, a_{2} \in \mathrm{A}$.

But this (from this website: Chapter 3) states defination of ......:

Given a set $A$ and a binary relation $R$ in $A, R$ is symmetric if and only if for every ordered pair $\langle x, y\rangle$ in $R,$ the pair $\langle y, x\rangle$ is also in $R .$ It is important to note that this definition does not require every ordered pair of $A \times A$ to be in $R .$ Rather for a relation $R$ to be symmetric it must always be the case that if an ordered pair is in $R,$ then the pair with the members reversed is also in $R$

Clearly both defination are contradicting each other(if they are not then please explain me.)

Let me explain the question by a example.

Let we a set $Z$ such that: $$ Z=\{1,2,3\} $$ And also let a set $P$ be such that: $$ P=\{(2,2)\} $$

Now according to first defination $P$ is not a symmetric relation on $Z$, 'cause $(3,1)\not=P$.

But $P$ is a symmetric relation on $Z$, according to second defination, because $$ P=\{(2,2) (2,2)\}$$. Which satisfies the condition of second defination.

So which one is correct?

Answer 1

• You are misled by the " for all" in the first definition, which you interpret as the requirement that the relation contains all the possible ordered pairs of $A\times A$ ( i.e. the cartesian product of set A by itself).

• The expression "for all $a_1, a_2 \in A$ " is simply a requirement regarding the origin of the ordered pairs : the ordered pairs must all come from $A\times A$. The expression "for all $a_1, a_2 \in A$ " is simply another way to recall that R is a relation on set A.

• The definition says :

for all pair $(a,b)_{\in A\times A} $ , IF $(a,b)$ is in R, THEN $(b,a)$ is also in R.

Answer 2

Each definition is equivalent to:

$\forall (a_1,a_2) \in A \times A : \lnot \left((a_1,a_2) \in R \land (a_2,a_1) \notin R\right)$

If we have $(a_1,a_2)$ with $(a_1,a_2) \notin R$ then $(a_1,a_2) \in R$ is false, so $(a_1,a_2) \in R \land (a_2,a_1) \notin R$ is also false, as required.