> [!definition] > > Let $E$ be a [[Locally Convex Topological Vector Space|locally convex topological vector space]], then the following are equivalent: > 1. $E$ is [[First Countable|first countable]]. > 2. There exists a countable family of [[Seminorm|seminorms]] $\seq{[\cdot]_n}$ that define the topology on $E$. > 3. There exists a translation-invariant [[Pseudo-metric|pseudo-metric]] on $E$ inducing its topology. > > If $E$ is [[Hausdorff Space|Hausdorff]], then the pseudo-metric is a [[Metric Space|metric]]. In which case, $E$ is complete as a topological vector space if and only if $E$ is complete as a metric space. > > If all of the above holds, then $E$ is a **Fréchet space**. > > *Proof*. If $E$ is first countable, then $E$ admits a countable neighbourhood base at $0$ consisting of [[Convexity|convex]], [[Balanced Set|balanced]], and [[Absorbing|absorbing]] sets. The [[Gauge|Minkowski functionals]] of each set in this base is a countable family of seminorms defining the topology on $E$. > > Suppose that there exists a countable family of seminorms $\seq{[\cdot]_n}$ that define the topology on $E$. Define > $ > d(x, y) = \sum_{n \in \nat}\frac{2^{-n}[x - y]_n}{1 + [x - y]_n} > $ > then $d$ is the desired pseudo-metric. > [!definition] > > Let $E$ be a Fréchet space, then $E$ is [[Bornologic Space|bornologic]]. > > *Proof*. Since $E$ is a Fréchet space, its metric is a [[Quasi-Norm|quasi-norm]]. For any seminorm $[\cdot]$ that is bounded on bounded sets, the identity map $(E, d) \to (E, [\cdot])$ is continuous. Therefore $[\cdot]$ is continuous on $E$.