Given a set $\{1,2,3,4\}$, how is the following relation $R$ antisymmetric?
$$R = \{(1, 2), (2, 3), (3, 4)\}$$
Note: Antisymmetric is the idea that if $(a,b)$ is in $R$ and $(b,a)$ is in $R$, then $a = b$. In my textbook it says the above is antisymmetric which isn't the case as whenever $(a,b)$ is in $R$, $(b,a)$ is not.
Try this: consider a relation to be antisymmetric, UNLESS there exists a counterexample: unless there exists $(a, b) \in R$ and $(b, a) \in R$, AND $a\ne b$.
Since no such counterexample exists in for your relation, it is trivially true that the relation is antisymmetric.
Another way to put this is as follows: a relation is NOT antisymmetric IF AND ONLY IF there exist $a, b$ such that BOTH $\;(a, b)\in R\;$ AND $\;(b, a) \in R\;$ BUT $\;a\ne b$.
This is true of other properties as well: a property holds for a relation unless there exists a counterexample such that the property fails to hold. Put differently, a property FAILS to hold IF AND ONLY IF a counterexample exists.
$R$ is antisymmetric iff whenever both $(a,b)$ and $(b,a)$ are in $R$ then $a=b$.
In your example, there is no pair $(a,b) \in R$ that also has $(b,a) \in R$, so the statement is vacuously true.
Another (equivalent) way of looking at it is that $R$ is not antisymmetric iff there are elements $a,b$ with $a\neq b$ and both $(a,b),(b,a) \in R$.
In my opinion your misunderstanding is in the logic, not in the set theory.
If $p$ is false, then the conditional statement “if $p$ then $q$” is vacuously true. This does not mean that $q$ is true, it means the entire statement is true.
As @Doug Chatham says,
For example, consider the implication, "If 2+2=5, then you will pass the course." Since 2+2 is not 5, the statement is a true statement, regardless of whether or not you pass the course.
Regarding your question, if $(a,b)\in R$ and $(b,a)\in R$ together imply $a=b$, and either one (or both) of $(a,b)\in R$ or $(b,a)\in R$ is false, then the entire statement is true: R is antisymmetric.
The definition of antisymmetry does not state
if $(a,b) \in R$ then $(b,a) \in R$
(that would be the definition of symmetry).
Here is an example that is going to show that the below relation is not anti-symmetric.
Let $R$ be the relation $$\bigl\{ (1,2), (1,3), (3,1), (1,1), (3,3), (3,2), (1,4), (4,2), (3,4)\bigr\} $$
First step is to find 2 members in the relation such that $(a,b) \in R$ and $(b,a) \in R$. If no such pair exist then your relation is anti-symmetric. If any such pair exist in your relation and $a \ne b$ then the relation is not anti-symmetric, otherwise it is anti-symmetric.
$$(1,3) \in R \text{ and } (3,1) \in R \text{ and } 1 \ne 3$$
therefore the relation is not anti-symmetric.
You just need to check the cases. You are given a set $A=\{1,2,3,4\}$ and the relation $$\sim\; =\{(1,2),(2,3),(3,4)\}$$
Note that $a\sim b\iff a+1=b$. Thus, it will be never the case that the other pair you're looking for is in $\sim$, and the relation will be antisymmetric because it can't not be antisymmetric, i.e. the truth holds vacuously.
