A. P. Robertson and W. Robertson in their ["Topological Vector Spaces"](https://rads.stackoverflow.com/amzn/click/com/0521298822) VII, 1.4, (and H.Jarchow in "Locally convex spaces", 4.6, Theorem 2) prove the following proposition:



> 
> Let $E=\lim\_{n\to\infty}E\_n$ be an inductive limit of a **sequence** of locally convex spaces $\{E\_n;\ n\in {\mathbb N}\}$, and suppose that for each $n\in{\mathbb N}$ 
> 
> 
> 1) the topology of $E\_n$ is induced from $E\_{n+1}$ (in this case $E$ is called a strict inductive limit of $E\_n$), and 
> 
> 
> 2) $E\_n$ is closed in $E\_{n+1}$. 
> 
> 
> Then each bounded set $B$ in $E$ is contained in some $E\_n$.
> 
> 
> 


I wonder if the following modification of this statement is true:



> 
> Let $E=\lim\_{\nu\to\infty}E\_\nu$ be an inductive limit of a net of locally convex spaces $\{E\_\nu;\ \nu\in I\}$ (where $I$ is directed, but not necessarily countable), and suppose that for each $\mu\le\nu\in I$ 
> 
> 
> 1) the topology of $E\_\mu$ is induced from $E\_\nu$, and 
> 
> 
> 2) $E\_\mu$ is closed in $E\_\nu$. 
> 
> 
> Then each [**totally bounded**](https://en.wikipedia.org/wiki/Totally_bounded_space) set $B$ in $E$ is contained in some $E\_\nu$.
> 
> 
> 


Or perhaps some other modifications hold (where "bounded" is replaced by "totally bounded" and the other conditions can be weakened instead)? 



When is a totally bounded set of an inductive limit contained in a component of this limit?
No. There are only very few general results about uncountable inductive limits. The following example is stated (without proof) in an article of Komura [Some examples on linear topological spaces. Math. Ann. 153 (1964), 150–162]: 


For an uncountable set $I$ and every countable $J\subseteq I$ let $E\_J=\{f:I\to \mathbb R: $supp$(f)\subseteq J\}$ (where the support is just $\{i\in I: f(i)\neq 0\}$) endowed with the Frechet topology of pointwise convergence ($E\_J$ is isomorphic to $\mathbb R^J$ with the product topology). Then $E=\lim\limits\_\to E\_J$ is a strict inductive limit and the limit topology is the relative topology of $\mathbb R^I$ (every neighbourhood of $0$ in $E$ contains the absolutely convex hull of $\bigcup\_J U\_J$ with $0$-neighbourhoods $U\_J$ of in $E\_J$ which only give conditions on values $f(j)$ for $j\in F(J)$ *finite*, then there is a finite subset $F$ of $I$ with $I\setminus F \subseteq \bigcup\_J J\setminus F(J)$).


Now $K=\{\delta\_x: x\in I\}$ is precompact in $E$ (because it is contained in $E$ and precompact in $\mathbb R^I$) but not contained in any step $E\_J$.


EDIT. A related example (certainly folklore, but I don't know about a reference): Again $I$ is an uncountable set and for each countable $J\subseteq I$ we set $E\_J=\ell^1(J)$ considered as a subspace of $\ell^1(I)=\{(x\_i)\_{i\in I}\in\mathbb R^I: \sum\_{i\in I}|x\_i|<\infty\}$ (where components outside $J$ are $0$). Then $E\_J$ is a strict inductive spectrum of Banach spaces and $E=\lim\limits\_\to E\_J=\ell^1(I)$ *topologically* (the identity $E\to \ell^1(I)$ is continuous because of the universal property of inductive limits, and the identity $\ell^1(I)\to E$ is sequentially continuous because every sequence in $\ell^1(I)$ is contained in a single $E\_J$, since $\ell^1(I)$ is metrizable, sequential continuity implies continuity). The unit ball of $\ell^1(I)$ is bounded but not contained iny step. The precompact sets however are indeed contained in a step. 


EDIT II


Let me try to make the coincidence of the product topology from $\mathbb R^I$ and the inductive limit topology a bit clearer: We have to estimate every continuous semonorm $p$ on $E$ by $cq\_F$ with a constant $c$ and $q\_F(f)=\max\{|f(j)|: j\in F\}$ for some finite set $F$ (these are the standard seminorms on the product). By definition of the inductive limit the restriction of $p$ to each $E\_J$ can be estimated by $c\_J q\_{F(J)}$ with some finite set $F(J)\subseteq J$, in particular, $p(f)=0$ for every $f\in E\_J$ with supp$(f)\cap F(J)=\emptyset$. 


I claim that for the family $(F(J))\_{J \text{ countable}}$ there is a finite set $F$ such that $I\setminus F \subseteq \bigcup\_J J\setminus F(J)$. Otherwise we find a countable and infinite subset $K$ of $I$ which is not in that union which contradicts the fact that $K\setminus F(K)$ is trivially contained there.


Next, let us see that we have $p(f)=0$ for every $f\in E$ with supp$(f)\cap F=\emptyset$. Let $K$ be the support of $f$. For each $k\in F(K)$ there is $J\_k$ such that $k\in J\_k\setminus F(J\_k)$. Using this we can write $f=g+\sum\_{k\in F(K)} f\_k$ such that $g\in E\_K$ vanishes on $F(K)$ and $f\_k\in E\_{J\_k}$ vanishes on $F(J\_k)$ which together with the subadditivity of $p$ yields $p(f)=0$.


Finally, since $E\_F$ is finite dimensional we find a constant $c$ such that $p(f)\le c q\_F(f)$ for $f\in E\_F$ which yields (again by decomposing $f\in E$ as $f=g+h$ with $g\in E\_F$ and $h$ vanishing on $F$) that $p(f)\le c q\_F(f)$ for all $f\in E$. 


