Let $\{X,\{X_i,T_i\}_{i\in I}\}$ be an inductive system of TVS's and let $T$ be the inductive limit topology. I am trying to show that the relative topology on $X_i$ induced by $T$ is coarser than $T_i$. This is where I've gotten so far:
$W\in T|_{X_i}$ implies $W=V\cap X_i$ for some $V\in T$.
But for all $v\in V$ there exists $U_v$, a balanced convex subset of $X$ such that $U_v\cap X_j\in T_j$ for all $j\in J$, and $(U_v+v)\subset V$.
Therefore $V=\cup_{v\in V}(U_v+v)$ and so $W=\cup_{v\in V}[X_i\cap(U_v +v)]$.
I would like to show that each $X_i\cap(U_v +v)$ is in $T_i$, as then the proof is finished, but not sure if this is true as I can't seem to find a way. Is my approach wrong?
I think this way works;
$B=\{U\in X : U\text{ convex, balanced and } U\cap X_i\in T_i \text{ for all } i\}$ defines a basis for the $(X,T)$ neighbourhood system at $0$.
Then $A=\{U\cap X_i : U\in B\}$ defines a basis for the $(X_i,T|_{X_i})$ neighbourhood system at $0$.
It's clear that $V\in A \implies V\in T_i$.
Since the neighbourhood system at $0$ defines the topology for a TVS, this is sufficient to show that $T|_{X_i}\subset T_i$.