It's my first post on this site, so please forgive me for all mistakes I've made during writing it. I can't understand the proof of this lemma:
Let $( X,\mathcal U )$ be an uniform space, $x\in X$ and $W,U \in \mathcal U $. If there exists $V\in \mathcal U$ such as $ U \circ V \subseteq W $ then:
- $U[x] \subseteq \operatorname{int} W[x]$
- $\overline{U[x]} \subseteq W[x]$
Proof:
- Let $y \in U[x] $ and $z\in V[y] $, then $(x,z) \in U \circ V $ , so $ (x,z) \in W $. It implies that $U[x] \subseteq\operatorname{int} W[x]$ ( I can't see how )
Let $y \in \overline{U[x]} $. Then for every $ Z \in \mathcal U $ exists $x_z \in U[x] \cap Z[y] $ ( I have no idea why this happens ) Then we choose $Z$ such that $Z=Z^{-1}$, $Z \subseteq V $. Now $(x,y) \in U \circ Z \subseteq W$.
Please, help me ;) If I've made language errors, I'd be very happy to hear about them.
The first one says that for all $y \in U[x]$, there is a neighbourhood of $y$ (namely $V[y]$) that is contained in $W[x]$. That means $y$ is an interior point of $W[x]$. Since that holds for all $y \in U[x]$, we have $U[x] \subseteq \operatorname{int}(W[x])$.
For the second one, $y \in \overline{U[x]}$ means that every neighbourhood of $y$ intersects $U[x]$. By the definition of the topology induced by a uniform structure, the sets $Z[y]$ for $Z \in \mathcal{U}$ are a neighbourhood basis for $y$. We could have done it simpler, if $y \in \overline{U[x]}$, then $U[x]\cap V^{-1}[y] \neq \varnothing$, since $V^{-1}[y]$ is a neighbourhood of $y$. Hence $y \in (U\circ V)[x] \subseteq W[x]$.