Definition 8.2.2 (Topology Induced by Pseudonorm). Let $E$ be a vector space over $K \in \RC$ and $\seqi{\rho}$ be pseudonorms on $E$. For each $i \in I$, let
then $d_{i}$ is a pseudometric. The uniform topology on $E$ induced by $\seqi{d}$ is the topology induced by $\seqi{\rho}$, and
The topology induced by $\seqi{\rho}$ is a vector space topology.
The uniformity induced by $\seqi{\rho}$ is the translation-invariant uniformity for its topology.
For each $i \in I$, $x \in E$, and $r > 0$, let $B_{i}(x, r) = \bracs{y \in E|d_i(x, y) < r}$, then
\[\bracs{\bigcap_{j \in J}B_j(0, r)|J \subset I \text{ finite}, r > 0}\]is a fundamental system of neighbourhoods at $0$.
Proof. (3): By Definition 5.3.3.
(2): Each $d_{i}$ is translation-invariant.
(1):
Let $x, x', y, y' \in E$, $J \subset I$ be finite, and $r > 0$. If for each $j \in J$, $d_{j}(x, x'), d_{j}(y, y') < r/2$, then $d_{j}(x + x', y + y') < r$ by (PN3).
Let $x, x' \in E$ and $\lambda, \lambda' \in K$, then
\[\lambda x - \lambda' x' = \lambda(x - x') + (\lambda - \lambda')x'\]Let $i \in I$ and $\eps > 0$. By (PN5) then there exists $\delta > 0$ such that if $\rho_{i}(x - x') < \delta$, then $\rho_{i}(\lambda (x - x')) < \eps$.
On the other hand,
\[(\lambda - \lambda')x' = (\lambda - \lambda')x + (\lambda - \lambda')(x' - x)\]By (PN4), there exists $\delta' \in (0, 1]$ such that if $\abs{\lambda - \lambda'}< \delta'$, then $\rho_{i}((\lambda - \lambda')x) < \eps$. In which case, since $\delta' \le 1$, (PN2) implies that
\[\rho_{i}((\lambda - \lambda')x') < \eps + \rho_{i}(x' - x) < 2\eps\]Therefore $\rho_{i}(\lambda x - \lambda' x') < 3\eps$.
$\square$