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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?

RakoonBerry
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1 Answers1

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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.

Nickel
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