I want to understand the quantifier for all ($\forall$) in the properties of a binary relation. Let $X=\{1,2,3\}$ and $R\subseteq X\times X$.
$R_1 = \{(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)\}$
$R_2=\{(1,2),(1,3),(2,3)\}$
$R_3=\{(1,1),(2,2),(3,3)\}$
$R_4=\{(1,1),(1,2),(1,3),(2,2),(3,3)\}$
The sets $R_1,R_3,R_4$ are reflexive, $R_3$ is symmetric and all of them are transitive. The book didn't say which one is antisymmetric. And my problem lies when I started to see why isn't some of them symmetric. From the property
$\forall x,y \in X$ $xRy$ $\implies$ $yRx$
I looked at the subset $R_3$ and started looking at the elements, when I see $\forall x,y$ I think about every possible combination of them ($X\times X$) who can satisfy that property:
$(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(2,3),(3,2),(3,3)$ so if this set was $R_5$ it would be symmetric, if I'm not wrong as it has all the possible combinations ?
So why is the set $R_3$ symmetric if it doesn't have all the combinations ($\forall x,y\in X$)
Look a little more carefully at the definition of symmetry of relation: some relation is symmetric if $\forall x, y \in X, xRy \rightarrow yRx$.
In phrasing this with $R \subset X \times X$, this means that such a relation is symmetric if anytime we have $(x,y) \in R$, we can also expect $(y,x) \in R$.
Under this phraseology, it is evident $R_3$ is symmetric, as for example if $(x,y) = (1,1)$, then $(y,x) = (1,1)$ and so forth for the other elements. Moreover, you can see here that your provided $R_5$ is also symmetric.
I think your confusion may stem from the quantification here, to say a relation is symmetric is not to say it contains all elements of the form $(x,y)$ and $(y,x)$ for arbitrary $y, x \in X$, but rather that we can expect $(y,x)$ when we already have $(x,y)$ and vice versa.