I'm asking because I was reading on antisymmetry from this question

Antisymmetric Relations

I may very well just be confused, but the relation doesn't state that 1 does not correspond to 3 as well. At the same time though, I would assume that it would only be transitive if the relation were


any help would be appreciated.

  • 1
    $\begingroup$ You are correct that $(1,3)$ is missing for this relation to be transitive. $\endgroup$ – Arnaud Mortier Oct 29 at 14:01
  • $\begingroup$ As an aside, since you were reading on antisymmetry, I suggest taking a look at this post of mine which may help clarify things further. $\endgroup$ – JMoravitz Oct 29 at 14:06

The relation $\mathcal{R}=\{(1,2),(2,3)\}$ (note, this is the relation $\mathcal{R}$, not "$x\mathcal{R}y$" which is a statement not the relation as a whole) is not transitive because for it to have been transitive we would have required that for every possible choice of $x,y,z$ (possibly repeating), if we had $x\mathcal{R}y$ and $y\mathcal{R}z$ that we would also have needed $x\mathcal{R}z$.

Since $(1,2)\in\mathcal{R}$ and $(2,3)\in\mathcal{R}$ but $(1,3)\not\in\mathcal{R}$ the relation is not transitive.

"The relation doesn't state that 1 does not correspond to 3 as well." On the contrary, the relation is very specifically defined to include those things listed in it and only those things listed in it. Since $(1,3)$ is not listed in the definition of $\mathcal{R}$, that directly confirms that $1$ is not related to $3$.


You are correct. Since $1R2$ and $2R3$ are true but $1R3$ is not true, the relation is not transitive.


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