Let $E$, $F$ be two LF-spaces (more details see here or here), $\{E_m\}$, $\{F_n\}$ $(m, n = 1, 2 .... )$ two sequences of definition of $E$ and $F$, respectively. If $u$ is an isomorphism of $E$ into $F$ (for the TVS structures), I want to prove that to every $n \in \mathbb{N}$ there is $m\in \mathbb{N}$ such that $$u^{-1}(F_n) \subset E_m.$$
I thought of the following: Let $m \in \mathbb{N}$. So, $E_m \subset E$ and $$u(E_m) \subset u(E) \subset \bigcup_{n=1}^{\infty} F_n,$$ what implies, there exists $n \in \mathbb{N}$ such that $$u(E_m)=F_n \Rightarrow E_m=u^{-1}(F_n) \Rightarrow u^{-1}(F_n) \subset E_m.$$ I think this is wrong. Any suggestion?
It seem that you mean by LF-space a strict LF-space (à la Dieudonné-Schwartz, Grothendieck did not require strictness). Then $L=u(E)$ is complete (being isomorphic to $E$) and hence closed in $F$. Then $F_n\cap L$ are closed in $F_n$. Then Grothendieck's factorization theorem for $u^{-1}: L\cap F_n\to E$ implies that $u^{-1}(F_n\cap L)$ is contained in some $E_m$ and this gives your claim because $u^{-1}(A)=u^{-1}(A\cap u(E))$ holds for every map $u:E\to F$ and every set $A$ in the range.