I am reading a book Axioms and Set Theory - A first course in Set Theory by Robert Andr´e
The example no. 4 discussed on page no. 52 under the section Equivalence relations and order relations
Let $S = \{a, b, c, d\}$. Consider the relation $R3 = \{(a, a), (b, b), (c, c), (d, d), (a, b), (b, a), (b, c), (c, b)\}$
· We see that $R3$ is a reflexive relation on $S$.
· Since $R3$ contains both pairs $\{(a, b), (b, a)\}$ and $\{(b, c), (c, b)\}$, then $R3$ is a symmetric relation on $S$.
· Since $R3$ contains $\{(a, b), (b, a), (a, a)\}$ and $\{(b, c), (c, b), (b, b)\}$, then $R3$ is anti-symmetric.
· Since $R3$ contains $(a, a)$, then $R3$ is not asymmetric.
· Since $R3$ contains the triples $\{(a, b), (b, a), (a, a)\}$ and $\{(b, c), (c, b), (b, b)\}$, then $R3$ is transitive on $S$.
Also on footnote it is mentioned the following statement:
Note that the statement “whenever $(a, b)$ and $(b, a)$ are in $S$, then $a = b$” holds true.
I understand that that $R3$ is symmetric and reflexive but I am not able to understand why $R3$ is anti-symmetric.
A relation $R$ is antisymmetric if $xRy\wedge yRx$ implies that $x=y$.
That is not the case for the relation $R3$ described in your question (provided that $a,b$ are distinct).
This because e.g. $(a,b)$ and $(b,a)$ both are elements of the relation while $a\neq b$.
So $R3$ is not antisymmetric.