The interior of the elements of a uniform structure $D$ is also base for $D$ The interior of the elements of a uniform structure $D$ is also base for $D$, i.e,
$B=\{V^\circ : V\in D\}$ is a base for $D$.
The proof is in my textbook, but I do not understand some parts.
Let us take $U \in D$. Then, there exists symmetric $V\in D$ st. $V^{3}\subset U$ (WHY?)
For all $(x,y)\in V$, $V(x)=\{z\in X : (x,z)\in V\}$ and $V(y)=\{z'\in X: (y,z')\in V\}$,
$V(x)\times V(y)=\{(z,z')\in X\times X | (x,z),(y,z')\in V\}$
Since $V$ is symmetric, $(x,z)\in V \implies (z,x)\in V$. Then,
Since $(z,x)\in V$ and $(x,y)\in V$, we get $(z,y)\in V^{2}$.
Similarly, $(z,y)\in V^{2}$ and $(y,z')\in V \implies (z,z')\in V^{3}$.
Therefore, $V\subset V^{3} \subset U \implies V\subset U^\circ$. (WHY?)
Since $U^\circ\subset U$ and $U^\circ\in D$, the proof is done.(WHY?)
 A: So, what is happening here?
For all entourages $X\times X\,\supseteq\, V\in\def\D{\mathcal D}\D\ $ we consider its interior $V^\circ$, according to the product topology of $X\times X$.


*

*By definition, any entourage is subdivisable, i.e. for any $U\in\D$ there is a $V_0\in\D$ such that ${V_0}^2\subseteq U$. 
Iterating this just once more gives a $V_1\in\D$ such that ${V_1}^2\subseteq V_0$, so in turn, ${V_1}^4\subseteq U$. 
Now, any entourage is reflexive, hence ${V_1}^3\subseteq {V_1}^4$. 
If we want a symmetric divident of $U$, first consider the symmetrization of $U$, that is, $U_1:=U\cap U^{op}\,\in\D$, then choose $V_0$ then $V_1$ as above for $U_1$, then symmetrize it, setting $V:=V_1\cap {V_1}^{op}$.

*Now $V(x)\times V(y)$ is a neighborhood of $(x,y)\in X\times X$, call it the '$V$-neighborhood'. 
We want to prove that those $(x,y)$ whose $V$-neighborhood is fully contained in $U\subseteq X\times X$, constitute an entourage. All such points lie in $U^\circ$, so it will imply $U^\circ\in\D$. 
Moreover, we find even a smaller set for that purpose, and that is $V$ itself, meaning $V\subseteq U^\circ$. 
So, the aim is to prove that $(x,y)\in V\,\implies\,V(x)\times V(y)\subseteq U$. 
For that end, $(z,z')\in V(x)\times V(y)$ will imply $\ z\,V\,x\,V\,y\,V\,z'\ $ so that $(z,z')\in V^3\ \subseteq U$.

*And then we are ready: We see that $U\in\D\implies U^\circ\in\D$, but the other direction also holds as $U^\circ\subseteq U$, and all entourages are filters. 
By the way, this is just the definition of base of a uniform structure: the upward closure of $\mathcal B$ will give back just $\D$.

