A uniform structure $\Phi$ on a set $X$ is a collection of relations that satisfy some special properties.
- In general, all relations must be reflexive.
- For metric spaces all the relations are symmetric too.
Also, the axiom $\forall U\in\Phi \exists V\in\Phi : V\circ V\subseteq U$ quite closer to being transitive, but not exactly that.
I am wondering if there is some uniform structure in which all the relations are transitive. i.e., $\forall U\in\Phi\rightarrow U\circ U\subseteq U.$
- If not, why?
- If yes, what would be the consequences of this special property? And can I see an example?
First of all, your claim concerning metric spaces is wrong. Let $\cal U$ be the uniformity associated with a metric space $X$. If $U$ is a relation of $\cal U$, then $U^{-1}$ is in $\cal U$. Thus the symmetric relation $U \cap U^{-1}$ also belongs to $\cal U$. Consequently, the uniformity $\cal U$ is generated by the symmetric relations of $\cal U$, but there are in general non-symmetric relations in $\cal U$.
Coming back to your question, suppose that all the relations in $\cal U$ are transitive and suppose that $X$ contains three distinct elements $a, b, c$. Let
$$ U = \{(x,y) \in X \times X \mid d(x,y) < d(a,c)/2\} \text{ and } V = U \cup \{(a,b), (b,c)\} $$ Then $U$ is an entourage and so is $V$, since it contains $U$. Now, as $V$ is transitive, one also has $(a, c) \in V$ and hence $(a, c) \in U$, a contradiction. It follows that $X$ contains at most two elements !
A uniformity is called transitive if it is generated by transitive entourages. This is an interesting notion, which, as mentioned in the comments, is the uniform analog of an ultrametric, like the $p$-adic metric.