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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.

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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.

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

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