Definition 8.10.1 (Inductive Topology). Let $\seqi{E}$ be TVSs over $K \in \RC$, $\seqi{T}$ such that $T_{i} \in \hom(E_{i}; E)$ for all $i \in I$, and $E$ be a vector space over $K$, then there exists a topology $\topo$ on $E$ such that:
$(E, \topo)$ is a TVS over $K$.
For each $i \in I$, $T_{i} \in L(E_{i}; E)$.
For any topology $\mathcal{S}$ on $E$ satisfying (1) and (2), $\mathcal{S}\subset T$.
For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T_{i} \in L(E_{i}; F)$ for all $i \in I$.
The topology $\topo$ is the inductive topology on $E$ induced by $\seqi{T}$.
Proof. (1): Let
To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply Proposition 8.1.12.
Every set in $\mathcal{B}$ is radial and circled by definition.
For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
Let $\topo$ be the vector space topology such that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, then $(E, \topo)$ is a TVS.
(2): For each $i \in I$ and $U \in \mathcal{B}$, $T_{i}^{-1}(U) \in \cn_{E_i}(0)$, so $T_{i} \in L(E_{i}; E)$.
(U): Let $U \in \cn_{(E, \mathcal{S})}(0)$ be circled and radial, then by (2), $T_{i}^{-1}(U) \in \cn_{E_i}(0)$ for all $i \in I$. Thus the circled and radial neighbourhoods of $0$ in $(E, \mathcal{S})$ is a subset of $\mathcal{B}$.
(4): Let $U \in \cn_{F}(0)$ be circled and radial and $i \in I$. Since $T \circ E_{i} \in L(E_{i}; F)$, $T_{i}^{-1}(T^{-1}(U)) \in \cn_{E_i}(0)$, so $T^{-1}(U) \in \mathcal{B}\subset \cn_{E}(0)$.$\square$