# How to geometrically interpret the composition of entourages?

In my topology course, we have been working for quite some time now with uniform spaces. I understand the metric background and I can work abstractly with uniformities, so that is great.

However, it has been bothering me since the beginning that I can't interpret the composition of entourages geometrically. I have already stared hours and hours at $$V\circ U:=\{(x,y)\in X\times X\mid \exists z\in X\colon (x,z)\in U, (z,y)\in V\},$$ (composition of entourages $U$ and $V$ in a uniform space $(X, \mathcal{U})$), but I just can't spin my head around what this means concretely. Sketching simple cases in the Euclidean case only makes things worse, because I always get confused midway...

Of course, I understand that this is a generalisation of the composition of maps, but that doesn't make it any clearer to me. My professor always says that $U$ is kind of "blown up with a factor $V$" (or the other way around) and I understand indeed that in particular $U\subset U\circ U$, but that doesn't really satisfy me.

The composition of relations can be interpreted as an operation on closeness. Let me first consider the case of metric spaces. The triangular inequality tells you that if $d(x,z) \leqslant a$ and $d(z,y) \leqslant b$, then $d(x,y) \leqslant a + b$. In other words, if $x$ and $z$ are $a$-close and if $z$ and $y$ are $b$-close, then $x$ and $y$ are $(a + b)$-close.

The situation is similar in a uniform space, but in a more abstract setting, where composition of relations replaces addition. Let $U$ and $V$ be entourages. If $x$ and $z$ are $U$-close and if $z$ and $y$ are $V$-close, then $x$ and $y$ are $(V \circ U)$-close.

• I assume you mean $d(x,z)\leq a$? – sqtrat Mar 30 '16 at 18:47
• @sqtrat Thanks! Edited... – J.-E. Pin Mar 30 '16 at 20:19