Determine if the relation A = R and for x, y ∈ R is reflexive, symmetric, transitive, or antisymmetric:
x ∼ y ⇔ x > 5. I think the relation is reflexive and not symmetric and not antisymmetric. Not too sure if it's transitive or not. How would you solve this question and write the answer in a proper way?
EDIT: So I just thought about this a bit more and now I don't think it's reflexive. I had a misunderstanding of the definition. My reasoning now is that it's not reflexive because you can take an example like (1,1) which isn't a part of A. It's not symmetric because (6,5) is a part of A but not (5,6). It's not antisymmetric because x=y is not always the case (e.g. (6,7)). Still not too sure if it's transitive or not. Can someone please correct me if I'm wrong?
At first you have to notice, that there is no condition for $y$ in $\sim$ therefore any $y \in R$ can be in the relation, since $\sim \subseteq R \times R$. So you decide whether $ x \sim y$ just by the value of $x$.
Let's look at the definition of the terms reflexive, transistive, andsymmetric:
reflexive :$\Leftrightarrow$ For all $x \in R$: $x \sim x$. This means for all $x$, without any exceptions. But $(1,1)$ is not in the relation since $x = 6 \gt 5$.
symmetric:$\Leftrightarrow$ For all $x,y \in R$: $x \sim y \Rightarrow y \sim x$. This is clearly not the case, because $6 \sim 1$ but not $1 \sim 6$.
transistive:$\Leftrightarrow$ For all $x,y,z \in R$ follows $(x \sim y \land y \sim z) \Rightarrow x \sim z$. This is true, since the $y$ must be greater than $5$ in order to fullfil the condition and make the implication true. And $z$ can be any variable as we have already shown.
Hint: make a sketch with all points in this relation.