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I understand that a relation is transitive if $(a,b), (b,c)$ and $(a,c)$ are elements of the relation $X$.

However, I don't understand why the following relation is transitive?

$X = \{(1,2),(1,3),(1,4)\}$

I know that this relation is anti-symmetric.

Any help is much appreciated, thanks.

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  • $\begingroup$ I was just typing a comment about the set notation missing, asking if that was intentional. Thanks, Harsh. $\endgroup$ – The Count Jan 3 '17 at 17:51
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    $\begingroup$ See vacuous truth en.wikipedia.org/wiki/Vacuous_truth $\endgroup$ – eepperly16 Jan 3 '17 at 17:52
  • $\begingroup$ @Peter I imagine he meant $R$ instead of $X$ to be equal to the list given. $\endgroup$ – TastyRomeo Jan 3 '17 at 17:55
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    $\begingroup$ @Peter a relation $R$ on a set $X$ is defined as any subset of the cartesian product $X \times X$. $\endgroup$ – TastyRomeo Jan 3 '17 at 17:57
  • $\begingroup$ Now I understand eepperly's comment. We do not have pairs $(a,b)$ and $(b,c)$ such that $a$~$b$ and $b$~$c$, therefore the transitivity is true by default. Thank you, steamyroot and amwhy! $\endgroup$ – Peter Jan 3 '17 at 18:00
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A relation $R$ is transitive if for every couple of pairs $\{ (x,y),(y,z)\} \subseteq R$ (i.e. second element of first pair is first element of second pair), you also have that $(x,z) \in R$.

In your case, no such couple of pairs exists, so no pair of the form $(x,z)$ must be in $R$ for it to be transitive!

To make this more formal, the definition of transitivity is $$\forall x,y,z \in X: \left((x,y) \in R \wedge (y,z) \in R\right) \implies (x,z) \in R$$ But, in your example, $\left((x,y) \in R \wedge (y,z) \in R\right)$ is never satisfied. Hence the implication is always true!

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  • $\begingroup$ I understand now, thank you so much! $\endgroup$ – nadrxj Jan 3 '17 at 17:56

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