In Direct limit, Martin rightly pointed out that my naive construction (now deleted) of the colimit (direct limit) of topological abelian groups was wrong. He shows how to do it properly (at least the coproduct) here.
Since then, I've been lurking some of the literature about the subject and this problem of the colimit of topological groups (about which I had previously no idea) seems at the same time classical and topical. For instance, this 1998's paper points out that the Encyclopedic Dictionary of Mathematics, second edition, MIT (1987), article 210, has made the same mistake that I did, stating that the direct limit of topological groups, with the inductive limit of topologies (my naive attempt) has always a continuous multiplication.
The authors of this paper show a counter-example (example 1.2, page 553) and here is my question: I must be absolutely dumb, but don't understand it. Could anyone help me?
For those who don't have access to the paper, here is the example.
Let $G_n = \mathbb{Q} \times \mathbb{R}^n$ with the usual topology. Imbed $G_n \hookrightarrow G_{n+1}$ as $x \mapsto (x,0)$. Then, as a plain abelian group, $G = \varinjlim_{n} G_n= \mathbb{Q} \times \prod'\mathbb{R}$, where $\prod'\mathbb{R}$ denotes the weak or restricted product, which is the way guys in this area call the direct sum; that is, elements of $\prod'\mathbb{R}$ are infinite tuples $( x_1, \dots , x_n, \dots )$, in which all of its components $x_n \in \mathbb{R}$ are zero, except a finite number of them.
The inductive limit topology is the finest one that makes all the inclusions $G_n \hookrightarrow G$ continuous. That is, $U \subset G$ is open if and only if $U\cap G_n$ is open for all $n$.
Let's see how they show that, with this inductive topology, the "product" (in fact, addition) $\mu : G \times G \longrightarrow G$ is not continuous ($\mu$ is the operation induced as an honest colimit of groups -no topologies- by the operations $\mu_n : G_n \times G_n \longrightarrow G_n$, which I assume are the usual additions of those linear spaces). In plain English:
$$
(x_0, x_1, \dots , x_n , \dots ) + (y_0, y_1, \dots , y_n , \dots ) = (x_0+y_0, x_1 + y_1 , \dots , x_n + y_n , \dots ) \ .
$$
So, in this situation it's enough to produce an open neighbourhood $U \subset G$ of the neutral element $e \in G$ such that $V^2$ is not in $U$, for any open neighbourhood $V$ of $e$ -where, I assume, $V^2$ means $V + V$.
Ok, here is the guy that is supposed to ruin (and sure it does) my naive attempt:
$$
U = \left\{
x = (x_0, x_1, \dots , x_n , \dots ) \quad \vert \quad \vert x_j\vert < \vert \cos (jx_0) \vert \ , 1 \leq j
\right\} \ .
$$
This guy is an open set of $G$ because $x_0$ being a rational number guarantees $\cos (jx_0) \neq 0$ for all $j$. Assume there is an open neighbourhood $V$ of $e$ such that $V^2 \subset U$. Then, $V \cap G_j$ contains an open interval $(-\varepsilon_j , \varepsilon_j)$ in $\mathbb{R}$ with $\epsilon_j > 0$ such that
$$
(-\varepsilon_0 , \varepsilon_0) \times (-\varepsilon_j , \varepsilon_j) \subset
\left\{
(x_0 , x_j) \in \mathbb{Q} \times \mathbb{R} \quad \vert \quad \vert x_j\vert < \vert \cos (jx_0) \vert
\right\} \ .
$$
And here come the two final sentences of the example I don't understand: This is impossible if $2j\varepsilon_0 > \pi$. A contradiction.
Any hints or remarks (even humiliating ones) will be welcome.
Best Answer
This means that $j\epsilon_0 > \frac{\pi}{2}$. Hence $-j\epsilon_0 < -\frac{\pi}{2}$. By the density of rationals, this tells you that there exists a sequence $q_1, q_2,\dots,q_n$ of rational numbers in $(-\epsilon_0,\epsilon_0)$ such that $jq_n\rightarrow \frac{\pi}{2}$. Hence $|\cos(jq_n)|\rightarrow 0$. Combining this with $|x_j| < |\cos(jq_n)|$ for all $n$ gives the contradiction.