Final topology of $\mathbb R^\infty$ with respect to $\mathbb R^n$/ weak topology I'm stuck on this example in Boto v. Querenburgs "Mengentheoretische Topologie" and I would really appreciate some insight from our more topologically savvy friends on here. =)
Let $I$ be an ordered set, and let $(X_j, \mathcal T_j)_{j \in I}$ be a family of topological spaces, such that for $j<k$, $j,k \in I$ we have:
$$X_j \subset X_k, \quad \mathcal T_j = \mathcal T_k|X_j$$
i.e. the topology on $X_j$ is induced by the injection $i_{jk}: X_j \hookrightarrow X_k$ from the topology on $X_k$.
On $X = \bigcup_{j\in I} X_j$ let $\mathcal T \; $  be the final topology with respect to $(i_j: X_j \hookrightarrow X)_{j\in I}$; This is called the weak topology on $X$.
Example: Let $X_n = \mathbb R^n$, and let $X = \mathbb R^\infty$.
Then a sequence $(x_k = (x_{k1}, x_{k2}, \dots ))_{k\in \mathbb N}$ converges to $x = (x_1, x_2, \dots )$ iff for any fixed $n$ the sequence $(x_{kn})_{k \in \mathbb N}$ converges to $x_n$.

Now I don't see why the last statement should be true. 
First off: To get a feel for this new kind of topology, I tried comparing it to other topologies on $\mathbb R^\infty$ known to me (and I think in the following already I must be making a mistake...)
Suppose $U = \prod_{n \in \mathbb N} U_n$ is open in the box topology on $\mathbb R^\infty$, i.e. $U_n \subset \mathbb R$ is open for all $n$. Then I think $U$ is also open in the weak topology: (I suppose $\mathbb R^j$ should be identified with $\mathbb R^j \times 0 \times 0 \times \dots \subset \mathbb R^\infty$, right?)
But then $i_j^{-1}(U) = \emptyset$ if $0 \notin U_n$ for some $n>j$ and $i_j^{-1}(U) = U_1 \times \dots \times U_j$ otherwise. Both of which are open in $\mathbb R^j$, thus $U$ should be open in the final topology w.r.t. the inclusions $i_j$.
Now consider the sequence 
$$x_j = \underset{\text{j-th component}}{(0, \dots, 0,\underbrace{1}, 0, 0, \dots)}$$
Clearly $x_j$ converges to $(0, 0,\dots)$ componentwise, but $x_j$ does not converge in the box topology, hence neither in the finer topology introduced in the example (the weak topology).

So where is the above argument wrong? What am I not understanding correctly about this topology?
Many, many thanks in advance for any useful comments and answers.
Regards,
S.L.
 A: I would think that the weak topology would be the smallest topology which induces the given topologies on $\mathbb R^n.$ Therefore since the product topology (with basis of the form $U_1\times\cdots\times U_n\times \mathbb R\times\mathbb R\times\cdots$) also induces the standard topology on the subspaces $\mathbb R^n$, it follows that the weak topology is at least as small as the product topology. Indeed, it is exactly the product topology in this case.
A: This is a situation in which some mildly categorical ideas are usefully clarifying, I think. Specifically, (in effect taking up a bit more aggressively some points already made in comments), the "inductive limit" or "colimit" of an ascending union of topological spaces (with continuous inclusions, of course, or things would be perverse) is (as they say in categorical situations) "unique up to unique isomorphism". One virtue of this is that there is no mandate to give a construction. The only possible issue (which is now-and-then delicate) is existence.
For locally convex topological vector spaces, it's not hard to prove that arbitrary colimits exist, including these ascending unions.
The usual construction exhibits colimits as quotients of coproducts (dually to the way (projective) limits are often subobjects of products). An interesting point is that in the category of locally convex t.v.s.'s, the coproduct (exercise-provably) has the diamond topology. This is subtly (and not to-me intuitively) different from the box topology. Certainly it is not the product topology, which is quite weak.
Perhaps some mild categorical considerations explain why some of the possible worries/questions here may be not-to-the-point.  
