On equivalence of defining sequences for inductive limits Let $E$ be a Hausdorff locally convex space and assume that $E = \mathrm{ind} E_n$ is the locally convex inductive limit of an increasing sequence of locally convex Hausdorff spaces $E_n \subseteq E$ and also that $E = \mathrm{ind} F_n$ is the locally convex inductive limit of an increasing sequence of locally convex Hausdorff spaces $F_n \subseteq E$. (The two defining sequences $E_n$ and $F_n$ are equivalent in the sense that they generate on $E$ the same topology.) Does it follow that for each $n$ there is a continuous injection $E_n \to F_m$?
Note that if all the spaces $E_n$ and $F_n$ are Fréchet (in which case $E$ is an (LF)-space) then this is true by Grothendiecks factorization theorem: any continuous linear map from a Baire space ($E_n$) to an (LF)-space ($E = \textrm{ind} F_n$) factors through a step ($F_m$).
 A: Not in general. We need some conditions on the spaces $E_n$ and $F_m$ for such a conclusion.
Consider $E$ the space of all complex sequences with finite support, and let $F_m$ be the subspace of sequences $x$ such that $k > m \implies x(k) = 0$. Since $F_m$ is finite-dimensional, it carries only one Hausdorff vector space topology, so there's nothing to choose. We endow $E$ with the locally convex inductive limit topology, which is the finest locally convex topology on $E$. We let $E_n$ be the subspace of $E$ with $k > n \implies x(2k+1) = 0$ for all $x\in E_n$, and endow it with the subspace topology induced by $E$. Then the locally convex inductive limit topology on $E$ is the finest locally convex topology such that all inclusions $E_n \hookrightarrow E$ are continuous, and since they are by definition continuous for the finest locally convex topology on $E$, the two sequences define the same inductive limit topology.
But since the $E_n$ are all infinite-dimensional, while the $F_m$ are finite-dimensional, no $E_n$ can be linearly injected into any $F_m$.
