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A binary relation $R$ over a set $X$ is transitive if whenever an element $a$ is related to an element $b$ and $b$ is related to an element $c$ then $a$ is also related to $c$.

If I consider any two ordered pairs there is no common element.So how is $R$ transitive in the above question?

Asaf Karagila
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Matix
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2 Answers2

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If $aRb$ and $bRc$ then $aRc$ this is the definition of transitivity. If there is no such $a,b,c$ then also statement is true . If word is important.

  • It's called vacuously true that a relation is transitive whenever the condition $aRb$ and $bRc$ is not satisfied. – amWhy Mar 27 '18 at 16:17
  • @amWhy and OP: I agree that if there is no such $a,b,c$, then the statement is vacuously true. But there are such $a,b,c$ in this problem - no where does it say $a,b,c$ are distinct. So, for example, $1R1$ and $1R1$ implies $1R1$ is true. – Jason DeVito - on hiatus Mar 27 '18 at 19:00
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Transitivity of $R$ means $\forall x,y,z\in A\;(\;((x,y)\in R\land (y,z)\in R)\implies (y,z)\in R).$

$\forall x,y,z\in A$ is an abbreviation for $\forall x\in A\;\forall y\in A\;\forall z\in A.$ It does NOT mean $\forall x\in A \;\forall y\in A\setminus \{x\}\; \forall z\in A. $ It does NOT exclude any of $x=y, y=z$ or $z=x.$

It is analogous to the definition of $A^3=A\times A\times A=\{(x,y,z): x,y,z\in A\}.$ That is $(x,y,z)\in A^3$ iff $(x\in A\land y\in A\land z \in A).$ If $x=y=z\in A$ then we certainly do have $(x\in A\land y\in A\land z\in A)$.

DanielWainfleet
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