Proof. For each $H \subset \natp$ finite, let

Define

then

1. Since $0 \in \bigcap_{H \subset \natp \text{ finite}}U_{H}$, $\rho(0) = 0$.

2. Let $x \in X$ and $\lambda \in K$ with $\abs{\lambda}\le 1$. By assumption (a), $\lambda U_{n} \subset U_{n}$ for each $n \in \natp$. Thus for any $H \subset \natp$ finite with $x \in U_{H}$,

   \[\lambda x \in \sum_{n \in H}\lambda U_{n} \subset \sum_{n \in H}U_{n}\]

   so $\rho(\lambda x) \le \rho(x)$.

3. Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_{M}$ and $y \in U_{N}$. Assume without loss of generality that $\rho_{M} + \rho_{N} < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_{P} = \rho_{M} + \rho_{N}$. In which case, $U_{P} \supset U_{M} + U_{N}$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.

For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1}< 2^{n}$, so $U_{n+1}\subset \rho^{-1}([0, 2^{-n}))$ by Proposition 3.1.5. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}}= U_{n}$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.

1. Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda}\ge \alpha$, $x \in \lambda U_{n}$. Therefore for any $\lambda \in K$ with $\abs{\lambda}\le \alpha^{-1}$, $\lambda x \in U_{n}$, and $\rho(x) \le 2^{-n}$.

2. Let $\lambda \in K$ and $n \in \natp$. By assumption (b), there exists $m \in \nat$ such that $\lambda U_{n-m}\subset \sum_{j = 1}^{m} U_{n-m}^{j} \subset U_{n}$.

$\square$