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I am asked: Determine whether the relation X on the set Z is reflexive, symmetric, antisymmetric, and/or transitive, where $(a,b) ∈ X$ if and only if a = 1.

Apparently, the set is antisymmetric and transitive. I understand how it is not reflexive or symmetric, but I don't get why it is transitive. If the set is composed of ${(1,0),(1,1),(1,2),(1,3),...}$, where is the transitivity found?

Johny
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2 Answers2

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Let's say we have $(a, b) \in X$ and $(b, c) \in X$. Since $(a, b) \in X$, we have $a=1$. Since $a=1$, we have $(a, c) \in X$, proving transitivity.

The interesting thing here is that because we have $(b, c) \in X$, we have $b=1$. Therefore, transitivity only applies when $(a, b)=(1, 1)$.

Noble Mushtak
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  • I don't understand why "since a = 1, we have (a,c) ∈ X" proves transitivity. Could you elaborate? – Johny Dec 29 '16 at 18:48
  • @Johny Remember, transitivity means if we have $(a, b) \in X$ and $(b, c) \in X$ implies $(a, c) \in X$. I assumed the first two statements and showed that, because the first means $a=1$, we have $(a, c) \in X$. This proof means that for any $(a, b), (b, c) \in X$, we will also have $(a, c) \in X$. That's what transitivity is, so that proves transitivity. – Noble Mushtak Dec 29 '16 at 19:31
  • Here are some examples of transitivity: $$(1, 1), (1, 2) \in X \implies (1, 2) \in X$$ $$(1, 1), (1, 0) \in X \implies (1, 0) \in X$$ $$(1, 1), (1, 5) \in X \implies (1, 5) \in X$$ As you can see, the $(a, b)$ ordered pair always has to be $(1, 1)$ for the hypothesis of transitivity to apply, as noted in the second paragraph of my answer. – Noble Mushtak Dec 29 '16 at 19:34
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You want to show that if $a \sim b$ and $b \sim c$ then $a \sim c$, where I've written $x \sim y$ for the statement that $(x,y) \in X$.

If $a \sim b$, then $a=1$. Hence $a \sim c$ for all $c$. We didn't actually need to assume that $b \sim c$.

For completeness, I'll go over why it's antisymmetric and not reflexive.

It's antisymmetric: if $a \sim b$ then $a=1$, while the only way $b \sim a$ is if $b=1$. Hence if both $a \sim b$ and $b \sim a$, then $a=b=1$.

It's not reflexive: $1 \sim 2$ but $2 \not \sim 1$.

Patrick Stevens
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