Lemma 5.1.11. Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then the following are equivalent:
$(x, y) \in V \circ M \circ V$.
There exists $(p, q) \in M$ such that $(x, p) \in V$ and $(q, y) \in V$.
There exists $(p, q) \in M$ such that $x \in V(p)$ and $y \in V(q)$.
There exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q)$.
There exists $(p, q) \in M$ such that $p \in V(x)$ and $q \in V(y)$.
There exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
Proof. $(1) \Rightarrow (2)$: By definition of composition, there exists $q \in X$ such that $(x, q) \in M$ and $(q, y) \in V$. Similarly, there exists $p \in X$ such that $(x, p) \in V$, $(p, q) \in M$, and $(q, y) \in V$.
$(2) \Rightarrow (3)$: By symmetry, $(p, x) \in V$. Thus $p \in V(x)$ and $q \in V(y)$.
$(3) \Leftrightarrow (4)$, $(5) \Leftrightarrow (6)$: By definition of product.
$(3) \Rightarrow (5)$: By symmetry, $a \in V(b)$ if and only if $b \in V(a)$.
$(5) \Rightarrow (1)$: By symmetry, $p \in V(x)$ implies that $(x, p) \in V$, and $(q, y) \in V$.$\square$