Symmetric relation definition - why is this false?

Question

Can someone explain to me why the following statement is false, according to my study materials for discrete mathematics?

If a relation $R$ on a set $X$ is symmetric, then $x\,R\,y$ and $y\,R\,x$ for all $x,y\in X$.

How can this be false? I thought that if a relation is symmetric, then you need to have $xy$ and $yx$ in both directions, for all elements of a set.

This is the answer provided in the book:

False. The definition states that in order for $R$ to be symmetric on $X$, we have that if $x\,R\,y$ for $x, y \in X$ then $y\,R\,x$ also. This not imply that all $x$ and $y$ in $X$ are related under $R$.

What am I missing here?

Answer 1

The textbook is right.

For example: $A=\{1,2,3,4,5,6,7,8,9\}$ and $R_1=\{(1,1),(2,3),(3,2),(4,3),(3,4)\}$ and $R_2=\{(2,2),(3,3),(1,2),(4,1),(3,4)\}$

Here $R_1$ is symmetric since $\forall\ (x,y) \in R_1$ there exists $(y,x) \in R_1$ but $R_2$ is not since $\forall\ (x,y) \in R_2$ there does not exist $(y,x) \in R_2$.

The definition of symmetry does not involve the set $A$. It is as the book says and I have mentioned above.

Answer 2

Definition of a symmetric relation $R$: a relation $R$ on a set $X$ is said symmetric if for all $x,y\in X$ such that $xRy$, then $yRx$.

This statement says that IF $x$ and $y$ are related, then $y$ and $x$ are related.

The false statement is $xRy$ and $yRx$ for all $x,y\in X$.

This statement makes the assumption that $xRy$ has a meaning whatever $x,y\in X$, which is false in general.

Answer 3

That statement is not closed (since R and X are free in it), and it can take on either truth value.
In particular, the two relations $\: \{\} \:$ and $\: \{\hspace{-0.03 in}\langle 0,\hspace{-0.04 in}0\rangle \hspace{-0.03 in}\} \:$ on $\: \{0\} \:$ are both symmetric
(since for the implication in the definition of symmetry, either the left side will be
false or the right side will be true), but $\: \big[\{\hspace{-0.03 in}\langle 0,\hspace{-0.04 in}0\rangle \hspace{-0.03 in}\} \:$ on $\: \{0\} \:$ satisfies the statement$\big]$
and $\big[\{\} \:$ on $\: \{0\} \:$ does not, since $\; x = 0 = y \;$ will be a counterexample in this case$\big]$.


On the other hand, the related statements

"For all sets X, for all relations R on X, if R
is symmetric, then xRy and yRx for all x,y ∈ X".

"For all sets X, there exists a relation R on X such that
if R is symmetric, then xRy and yRx for all x,y ∈ X".

"There exists a set X such that for all relations R on X,
if R is symmetric, then xRy and yRx for all x,y ∈ X".

"There exists a set X and a relation R on X such that
if R is symmetric, then xRy and yRx for all x,y ∈ X".

are each sentences in the logical sense, and each of them has a well-defined truth value.

Answer 4

Let $X$ be the set $\mathbb{N}$ of natural numbers, and let $R$ be $=$, the usual equality relation. Then $=$ is symmetric, because $x=y$ and $y=x$ mean the same thing.

However it is not true that all natural numbers $x$ and $y$ fulfill $x=y$. For example $1=2$ is not satisfied.

If $x$ equals $y$, then $y$ equals $x$. But not all pairs $x,y$ are like that.