You did fine with reflexivity, and with symmetry and antisymmetry.
Now, let's look at transitivity:
We can summarize the relation as follows: $xRy$ if and only if $x$ and $y$ differ by $1$.
So, suppose $xRy$ ($x$ and $y$ differ by one) and $yRz$ ($y$ and $z$ differ by one),
What may be the case about the difference between $x$ and $z$?
- (Suspect a counterexample exists: just find $x, y, z$ such that $x = y - 1, y = z - 1 \implies x = z - 2$.
Or, vice versa, $x = y+1, y = z+1 \implies x = z+2)$
Let $x = 0$, $y = 1$, and $z = 2$, so we certainly have $x, y, z \in \mathbb Z$
- Clearly, $x = y - 1$ since $0 = 1-1$, so $x R y$,
- And $y = z - 1$, since $1 = 2 - 1$, so $y R z$.
- But it is not true that $x = z + 1 $, since $0 \neq 2+1 = 3$ nor
does $x = z - 1$, since $0 \neq 2 - 1 = 1$.
Hence, $x$ is not related to $z$, and transitivity fails.
All we need is one counterexample to prove that a relation is non-transitive, and we've just found one such couterexample
It is reflexive if this is a relation over the set $\{1,2,3,4\}$, and yes, the relation is symmetric.
Yes, If we remove $(1,2)$ or $(2,1)$ then it is anti-symmetric.
The relation is transitive, we do not need $(2,3)$ and $(3,4)$ to be in the set. Especially there is no pairs in the relation $(2,x)$ and $(x,3)$, which is what we would need in order to force $(2,3)$ to be in the relation due to transitivity.
Best Answer
Let $A = \{a, b, c\}$ and $R = \{(a, c), (b, b), (c, a)\}$ be a relation on $A$.
Reflexive?
We need to have that for all $x \in A$, $(x, x) \in R$.
Symmetric?
We need to have that for all $x, y \in A$, if $(x, y) \in A$ then $(y,x)\in A$.
Transitive?
We need to have that for all $x, y, z \in A$, if $(x, y)$ and $(y, z)$ are in $R$, then $(x, z)$ is in $R$.
Antisymmetric?
We need to have that for all $x, y \in A$, if $(x, y), (y, x) \in R$, then $x = y$.