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I know it is not complete and I know it is asymmetric. But how is it transitive? The textbook says it is transitive...

Instead, I used a different example of transitive, asymmetric, and not complete, which is $X = \{a, b, c, d\}$ with $R = \{(a, b), (b, c), (a, c)\}$.

PrincessEev
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    It is indeed transitive. The reason being, for any collection of elements, $x,y,z$ (some of these are allowed to be equal) whenever you have $xRy$ and $yRz$ you must also have $xRz$. In your specific example there do not exist any three elements $x,y,z$ even allowing repeats such that $xRy$ and $yRz$ simultaneously. As a result, it is vacuously transitive. This is the same reason why the relation ${(a,b)}$ is transitive. See vacuous truth. – JMoravitz Dec 11 '17 at 00:37
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    A different wording of transitivity in terms of graph theory: any directed path from point $a$ to point $b$ that can be made in two (or more) steps must also be traversable in a single step. – JMoravitz Dec 11 '17 at 00:42
  • Thank you, I always forget about vacuous truths...! –  Dec 11 '17 at 00:57
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    As an aside, if you are looking for the smallest example of a transitive antisymmetric incomplete relation, the empty relation on the set ${1}$ would suffice. – JMoravitz Dec 11 '17 at 01:03

2 Answers2

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As noted in the comments, this is indeed a transitive relation. This is because the definition of transitivity requires "$xRy \land yRz$" to hold; however, for the given relation, this never holds, and thus via a vacuous argument the relation is transitive.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.

PrincessEev
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Another view on this. Let $R\subseteq X\times X$. Define $R^2=\{(a,c)\,|\, \exists b\in X.(a,b),(b,c)\in R\}$.

By definition $"R \text{ is transitive}"\Leftrightarrow R^2\subseteq R$.

In your case (in the title) $R^2=\varnothing$.

For your second relation we have $R^2=\{(a,c)\}$.

MphLee
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