# When is a totally bounded set of an inductive limit contained in a component of this limit?

A. P. Robertson and W. Robertson in their "Topological Vector Spaces" 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 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)?

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 $$f\in E$$.

• Jochen, excuse me, I could not reply before. I don't understand this place: "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 $I\setminus F \subseteq \bigcup_J J\setminus F(J)$". As far as I understand, you mean that the topology of $E$ is induced from ${\mathbb R}^I$. I did not understand, why. – Sergei Akbarov Apr 5 '19 at 3:59