This question provided a geometric interpretation for transitivity in equivalence relations, but what about just transitivity by itself, without reflexive and symmetric constraints?
For example, reflexive relations can be visualised as relations that include the x=y line, symmetric as those that are reflected over said line, but what about transitive relations without those two properties?
From graphing out a few of them myself, it seems like they have to form some sort of right triangle, since if (x,y) and (y,z) are in the relation, so must (x,z), since two points share the same x coordinate and another two shares the same y. But is there a way of telling if a relation is transitive just by looking at all of the points plotted on a 2d plane? What does it even look like for all points to form right triangles with another two?
Best Answer
I cannot think of a way to visualize this with right triangles. But, perhaps there is another geometric way to think about it, using reflections across the $45^\circ$ line $y=x$. I don't know how useful this is, but here goes.
Let's fix $x=a$.
Draw the vertical line $x=a$.
Take all of the points $(a,b)$ on that line which are in the relation.
Draw all of the horizontal lines $y=b$.
Reflect those lines across the line $y=x$, and now we have a bunch of vertical lines $x=b$.
Take all of the points $(b,c)$ on those vertical lines which are in the relation.
Draw all of the horizontal lines $y=c$. Intersect them all with the vertical line $x=a$. All of those intersection points must also be in the relation.