# Is this relation, $xRy=\{(1,2),(2,3)\}$, transitive?

$$xRy=\{(1,2),(2,3)\}$$

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

$$xRy=\{(1,2),(2,3),(1,3)\}$$

any help would be appreciated.

• You are correct that $(1,3)$ is missing for this relation to be transitive. – Arnaud Mortier Oct 29 at 14:01
• 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. – 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.