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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.



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Why is the box topology coarser than the weak topology? Isn't the weak topology the coarsest topology which endows the spaces with the induced topology based on the inclusions (not the finest topology that makes the inclusions continuous)? –  Arturo Magidin Mar 23 '11 at 18:59
@Arturo: If my argument above were correct, it would show that any set open in the box topology is also open in the weak topology, hence the weak topology would have to be finer than the box topology. To your second remark: I only have the definition given above at hand. But are the two possibilities necessarily contradictory? –  Sam Mar 23 '11 at 19:40
@S.L. The two notions may yield different topologies: the product topology on $\mathbb{R}^{\infty}$ makes all inclusions $i_n\colon\mathbb{R}^{n}\to\mathbb{R}^{\infty}$ that you have defined continuous, and gives $\mathbb{R}^n$ the induced topology; so does the box topology. If you are looking for the coarsest topology that induces the topologies via the inclusions, then it will be coarser than the product topology; if you are looking for the finest topology that makes all inclusions continuous, then it would be finer than the box topology. So clearly, the two notions are not equivalent. –  Arturo Magidin Mar 23 '11 at 19:43
@S.L. Of course, the two notions may agree on some families: if the family has a maximum element, then both topologies (coarsest making all topologies induced by the inclusions, and finest making all inclusions continuous) would be the topology of the maximum. –  Arturo Magidin Mar 23 '11 at 19:45
@Arturo: Yes, you are right. And then with the definition as given (which is a one-to-one translation from the German text in front of me), you don't think there is necessarily something wrong with the argument above? (The example given in the book including the remark about convergent series then would have to be wrong with the definitions stated?) –  Sam Mar 23 '11 at 20:49
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2 Answers 2

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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.

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@Jim Conant: Doesn't the box topology also induce the same topology on the subspaces $\mathbb R^n$? It seems to me that this is the case, since any open set in the box topology whose intersection with $\mathbb R^n$ is nontrivial restricts to an open set of the form $U_1 \times \dots \times U_n$ on $\mathbb R^n$..? –  Sam Mar 23 '11 at 19:43
Yes, but it is not the smallest (coarsest) such topology to do so. –  Grumpy Parsnip Mar 23 '11 at 19:44
I agree with everything you say, but then you are implying the author simply gave the wrong definition of weak topology? –  Sam Mar 23 '11 at 20:51
@ S.L.: are you quoting him verbatim? As you can see, defining the topology to be consistent with the topologies on the $X_j$ spaces is not well-defined, since there is more than one such topology. To get a well-definition, you have to do something like taking the coarsest such topology. –  Grumpy Parsnip Mar 23 '11 at 21:00
I'm not really quoting him since it is in German :) , but the translation should be quite exact. His definition is well-defined, though: The final topology with respect to $(i_j: X_j \to X)_{j\in I}$ is exactly the finest topology, such that all inclusions $i_j$ are continuous. Which seems to be at the exact opposite end of the spectrum for a topology restricting nicely to the subspaces $X_j$ from what you and Arturo proposed. –  Sam Mar 23 '11 at 21:28
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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.

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