The neighbourhood filter of each point is a minimal Cauchy filter

Question

Wikipedia says (for every fixed uniform space) "The neighbourhood filter of each point (the filter consisting of all neighbourhoods of the point) is a minimal Cauchy filter."

Please help me to prove this statement.

Answer 1

Let $X$ be our uniform space, and consider $\mathscr{N} = \mathscr{V}(x)$, the neighbourhood filter of an arbitrary point $x\in X$. Let $\mathscr{C} \subset \mathscr{N}$ be a Cauchy filter. We need to show that $\mathscr{N} \subset \mathscr{C}$.

So let $U \in \mathscr{N}$ arbitrary. By definition of the topology there is an entourage $\mathscr{U}$ with $\mathscr{U}(x) = \{ y : (x,y) \in \mathscr{U}\} \subset U$. Since $\mathscr{C}$ is a Cauchy filter, there is a $C\in \mathscr{C}$ with $C\times C \subset \mathscr{U}$. Since $\mathscr{C}\subset \mathscr{N}$, $C$ is a neighbourhood of $x$, in particular, $x\in C$. Thus $\{x\}\times C \subset \mathscr{U}$, and hence $C \subset \mathscr{U}(x) \subset U$, whence $U \in \mathscr{C}$. $U$ was arbitrary, so $\mathscr{N}\subset \mathscr{C}$, showing the minimality of $\mathscr{N}$.