Proposition 8.1.6. Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then:
There exists a unique translation-invariant uniformity $\fU$ on $E$ that induces the topology on $E$.
For each neighbourhood $V \in \cn(0)$, let $U_{V} = \bracs{(x, y) \in E^2| x - y \in V}$, then for any fundamental system of neighbourhoods $\fB_{0}$ at $0$, $\fB = \bracs{U_V| V \in \fB_0}$ is a fundamental system of entourages for $\fU$.
The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
Proof. (2): Firstly, for any $V \in \fB_{0}$, $U_{V}$ is translation-invariant.
For any $V, V' \in \fB_{0}$, there exists $W \in \fB_{0}$ such that $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'}\supset U_{W} \in \fB$.
For any $V \in \fB_{0}$, $0 \in V$, so $\Delta \subset U_{V}$.
Let $V \in \fB_{0}$, then by (TVS1), there exists $W \in \fB_{0}$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_{W}$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_{W} \circ U_{W} \subset U_{V}$.
By Proposition 5.1.8, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
(1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
Let $W \in \cn(0)$, then by Lemma 8.1.5, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_{W}$. Thus $\mathfrak{V}\supset \fU$.
Let $V \in \mathfrak{V}$. Using Lemma 8.1.5, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_{0}$ with $W \subset V(0)$. In which case, $U_{W} \subset V$, and $\fU \supset \mathfrak{V}$.$\square$