Let $(X,\mathscr U)$ be a uniform space, i.e. $X$ is a set and $\mathscr U\subset 2^{X\times X}$ is a filter such that
- $\Delta\subset U$ for all $U\in\mathscr U$
- $U\circ V,U^{-1}\in\mathscr U$ for all $U,V\in\mathscr U$
- For all $U\in\mathscr U$, there exists some $V\in\mathscr U$ such that $V\circ V\subset U$
where we define $$U^{-1}=\{(y,x)\ |\ (x,y)\in U\}$$ and $$U\circ V = \{(x,z)\ |\ (x,y)\in U,(y,z)\in V\hbox{ for some }y\in X\}.$$ Furthermore we define $$U[x]=\{ y\in X\ |\ (x,y)\in X\}.$$
Meanwhile, for any set $X$ and family of filters $(\mathcal N_x)_{x\in X}$ in $X$, we say that $(\mathcal N_x)_{x\in X}$ forms a neighborhood filter base if
- $x\in N$ for all $N\in\mathcal N_x$
- For all $N\in\mathcal N_x$, there exists some $M\subset N$, $M\in\mathcal N_x$ such that $y\in M$ implies that $M\in\mathcal N_y$.
A neighborhood filter base thus defines a topology $\tau$ on $X$. Note that we still have a topology $\tau$ on $X$ if we remove condition 2, but in that case, $\mathcal N_x$ might not be the neighborhood filter associated with $\tau$ at $x$.
Now, if we set $\mathcal N_x=\{U[x]\ |\ U\in\mathscr U\}$ for all $x\in X$, then it is clear that $\mathcal N_x$ is a filter satisfying condition 1, but I have yet to succeed in proving condition 2. I suspect I must use conditions 2 and 3 for the definition of a uniform space, but I don't see how.