Proposition 2

Let $X$ be a uniform space. For every symmetric entourage $V$ of $X$ and every subset $M$ of $X \times X$, $VMV$ is a neighbourhood of $M$ in the product space $X \times X$, and the closure of $M$ in this space is given by the formula

where $\mathscr{S}$ denotes the set of symmetric entourages of $X$.

Let $V$ be a symmetric entourage of $X$. The relation $(x, y) \in VMV$ means that there is an element $(p, q)$ of $M$ such that $(x, p) \in V$ and $(q, y) \in V$: in other words ($V$ being symmetric) $x \in V(p)$ and $y \in V(q)$, that is $(x, y) \in V(p) \times V(q)$. Since $V(p) \times X(q)$ is a neighbourhood of $(p, q)$ in $X \times X$, the first part of the proposition is proved. The relations $(x, p) \in V, (y, q) \in V$ can also be written $p \in V(x), q \in V(y)$ or $(p, q) \in V(x) \times V(y)$. As $V$ runs through $\mathscr{S}$, the sets $V(x) \times V(y)$ form a fundamental system of neighbourhoods of $(x, y)$ in $X \times X$; for if $U, U'$ are any two entourages there is always a symmetric entourage $V \subset U \cap U'$, so that $V(x) \times V(y) \subset U(x) \times U'(y)$. Hence $V(x) \times V(y)$ meets $M$ for each $V \in \mathscr{S}$ if and only if $(x, y) \in \overline{M}$, and formula (1) follows.