I am trying to read about unifrom spaces from Introduction to Uniform Spaces, and I was wondering about some basic facts which I wasn't able to find there. Let $(X,\mathcal{E})$ be a uniform space.
- Can we find a base $\mathcal{B}$ for $\mathcal{E}$ such that $D\supseteq D^2$ for all $D\in \mathcal{B}$? I actually need to check whether we can say that $D[A]\supseteq D^2[A]$ for any set $A\subseteq X$.
- This may be a trivial question, but can we say that $E^2\supseteq E$ for any $E\in \mathcal{E}$? I think this follows from the fact that $E\supseteq \Delta X$, but I am not sure.
- If $\mathcal{E}$ is a left uniform structure on a topological group $G$, can we say that base entourages are of the form $U\times U$, where $U$ is a neighbourhood of $e\in G$?
Since I am trying to read this without prior experience, I would appreciate any relevant insights. Thanks to all responders.
The answer to the first is no, though some uniformities (the discrete one) this can indeed be done, taking $\{\Delta_X\}$ as that base. Maybe such a base implies that the uniformity is discrete, I haven't checked. But $D^2 \subseteq D$ is quite rare ( if $D$ were also symmetric, we’d have an equivalence relation as entourage)..
Of course (as to 2) we always have $E \subseteq E^2$ almost trivially: $(x,y) \in E$ then also $(y,y) \in E$ as $\Delta_X \subseteq E$ and so $(x,y) \in E^2$ by definition...
For 3 no: a standard base for the left uniformity is all sets of the form $$\bigcup_{g \in G} (gU \times gU)$$ where $U$ ranges over a neighbourhood base of $e$ in $G$. Such sets have at least the property that the contain $\Delta_G$, as they should, while $U \times U$ does not so cannot be an entourage at all.