I've been struggling with identifying relations on a set, and was hoping someone could check my answers and make sure I'm on the right track.
Let $A = \{1,2,3,4\}$ and $R$ be a relation on the set $A$ defined by:
$R = \{(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,2),(4,4)\}$
Determine whether $R$ is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive. Here are the answers that I came up with – do the answers (and my explanations) make sense at all? Thanks!
Reflexive, because of $(1,1), (2,2), (3,3), (4,4)$ (if the pairs were graphed they would point back to themselves).
NOT irreflexive because of $(1,1), (2,2), (3,3), (4,4)$ (because there are pairs that point back to themselves).
NOT symmetric because there is no $(4,1)$ or $(2,4)$ (if the pairs were graphed they would all need to connect in both directions).
NOT asymmetric because there is $(1,2)$ and $(2,1)$ (no pairs should connect in both directions if they were graphed).
ANTISYMMETRIC because of $(1,2)$ and $(2,1)$ (both are in the set, and $1$ does not equal $2$).
TRANSITIVE because of $(1,2), (2,1), (1,1), (2,2)$ or there’s a $(1,4)$ and a $(4,2)$ and also a $(1,2)$ (if there’s an $(a,b)$ and a $(b,c)$ there has to be an $(a,c)$).
Best Answer
You've done a fine job, though you're off on antisymmetry and also transitivity.
For antisymmetry: The reasoning you give is precisely why it is NOT antisymmetric.
If it were antisymmetric, then $(1, 2) \in R$ and $(2, 1) \in R$ would imply $1 = 2$, which is absurd.
With respect to transitivity: We see that $(4, 2), (2, 1) \in R$, but $(4, 1)$ is not in $R$. So the relation is not transitive.