If we graph equivalence relations on $\mathbb{R}$ on the plane $\mathbb{R} \times \mathbb{R}$, the properties of reflexivity and symmetry give rise to certain geometric properties--i.e. reflexivity means the line $y = x$ must be included, and symmetry means the graph must be symmetric about this line.
However, for transitivity, it being defined in terms of three different points, it's harder for me to pin down a visual interpretation in this regard. In some sense, the structures formed seem to be "square-like" (for example, the graph of the smallest equivalence relation containing the set $$S = \{(x,y)\,|\,(y = x + 1) \wedge (0 < x < 2)\},$$ this example being from Ch. 1, Section 3, Exercise 5 of Munkres' Topology, for reference).
In short, is the "square-like" interpretation true? How can this be formalized? If it's not true, what is a good way to visually interpret transitivity as a property of relations on $\mathbb{R}$ by graphing them in the plane, if such a way even exists?
The axiom explains itself, does it not?
For each $n$, and for each point $P$ on $x=n$, and point $Q$ on $y=n$, there must be a fourth point $Y$ in the relation such that $P-(n,n)-Q-Y$ is a rectangle. (You could say that if $P$ or $Q$ coincided with $(n,n)$, you'd be looking for a degenerate rectangle that's either a line or a point.)