If you have a uniform space say $X$ with a uniformity $U$, by definition there is an element of $U$ such that $E\circ E \subseteq D$ whenever $D\in U$. Apparently you are actually able to find a symmetric element satisfying that $E\circ E\circ E\subseteq D$.
Is this true? I have been trying to show it for a while but havn't been able to get it.
On
Every $U$ in the uniformity contains a symmetric uniformity: take $V = U \cap U^{-1}$. Clearly $V \subseteq U$ and $V$ is in the uniformity as these are closed under finite intersections and the inversion operation.
And $(x,y) \in V$ implies that $(x,y) \in U$ and $(x,y) \in U^{-1}$,so that $(y,x) \in U$ and $(y,x) \in U^{-1}$,so indeed $(y,x) \in V$. So $V$ is symmetric.
To go from triple composition from double is straightforward. From $D$ find $U$ with $U \circ U \subseteq D$. Then find $V$ such that $V \circ V \subseteq U$. Then $V \subseteq U$ (suppose $(x,y) \in V$, as $(y,y) \in V$ as well: $(x,y) \in V \circ V \subseteq U$). Then $V \circ V \circ V \subseteq D$:
Take $(x,y) \in V \circ V \circ V$ so that there exist $x_2, x_3 \in X$ such that $(x,x_2) \in V, (x_2,x_3) \in V \text{ and } (x_3, y) \in V$. The first two give $(x, x_3) \in V \circ V$ so $(x, x_3) \in U$, also $(x_3, y) \in U$ as well, so $(x, y) \in U \circ U \subseteq D$. The inclusion has been shown.
Now follow the first fact and define $E = V \cap V^{-1}$, which is symmetric and a subset of $V$ so $E \circ E \circ E \subseteq V \circ V \circ V \subseteq D$, as required.
Given $D$, first find $H$ such that $H\circ H\subseteq D$. Then find a symmetric $E$ such that $E\circ E\subseteq H$. (If your axioms give you an $E$ that isn't already symmetric, take $E\cap E^{-1}$ instead). We now have $$ E\circ E\circ E \;=\; \Delta \circ E \circ E \circ E \;\subseteq\; E\circ E\circ E\circ E \;\subseteq\; H \circ H \;\subseteq\; D $$