Is addition continuous?

Question

I'm going to ask a very silly question, so I'm begging you to be understanding if it is absolutely trivial, or if it's an exercise in some Bourbaki. I'm afraid of asking you, because the question entails this particular case: is the addition of (a finite, but variable quantity of) real numbers a continuous function?

Ok, let me try to explain what I mean.

I'm insisting in doing exercices about topological groups, see if I can understand something about those previous question 1, question 2 and question 3. This one should be more elementary.

Let $G$ be a topological Abelian group, written additively, and let $I$ be an arbitrary set. I don't mind if you take $G = \mathbb{R}$, with the usual Euclidian topology and the usual addition of real numbers, but you cannot assume $I$ is finite, nor countable. Let

$$ G^I = \prod G = \prod_\alpha G_\alpha , \qquad \text{where}\ G_\alpha = G \qquad \text{for all}\ \alpha \in I \ . $$

Let us denote $(x_\alpha)$ the elements of $\prod G$. This is an Abelian topological group with the usual product topology and operations defined component-wise:

$$ \begin{align} (x_\alpha) + (y_\alpha) &= (x_\alpha + y_\alpha) \\\ -(x_\alpha) &= (-x_\alpha) \end{align} $$

Consider also the weak product $\prod' G$ (aka, direct sum). That is, elements of $\prod' G$ are those tuples $(x_\alpha) \in \prod G$ such that $x_\alpha = 0$, except for a finite number of indexes $\alpha \in I$.

$\prod'G $ is of course a subgroup of $\prod G$, $\prod' G \subset \prod G$. Let's consider the subspace topology on $\prod' G$, induced from the product topology on $\prod G$.

Then we have a well-defined addition map that we hadn't in $\prod G$:

$$ \sum : \prod ' G \longrightarrow G \ , \qquad (x_\alpha ) \mapsto \sum_\alpha x_\alpha \ . $$

This makes sense, despite the fact $I$ is not necessarily an ordered set because, given an element $(x_\alpha) \in \prod'G$, if $x_{\alpha_1}, \dots , x_{\alpha_n}$ are its non-zero components, we don't have a canonical choice in order to perform the addition $x_{\alpha_1} + \cdots + x_{\alpha_n}$. Nevertheless, since the sum is associative and commutative, we may do it as we please: the result will always be the same. Hence $\sum_\alpha x_\alpha$ is well-defined.

Question. Is this map $\sum$ continuous?

That is, if you don't like too much abstraction in your life, but prefer to be very specific: is the sum of real numbers

$$ \sum: \prod' \mathbb{R} \longrightarrow \mathbb{R} \ , \qquad (x_\alpha) \mapsto \sum_\alpha x_\alpha $$

continuous?

Remark 1. The question seems silly, doesn't it? Well, maybe it is, but the first answer that came to my mind is wrong, as far as I can see: you cannot say "this $\sum$ is continuous because it is a composition of continuous maps"; namely, the iteration of the addition of $G$. Yeah, but: which composition? Notice that there is no canonical choice for doing the addition in an specific order -there could be no order at all in $I$. So every time you compute $\sum_\alpha x_\alpha$ you can change the order in which you perform the (continuous) sums $x_\alpha + x_\beta$. So, $\sum$ is not a composition of a finite number of specific maps.

(Edit. Previous Remark 2 was wrong.)

Answer 1

In the case where $I$ is infinite and $G$ has a non trivial topology, then the answer is no. let $(x_\alpha) \in \prod 'G_\alpha$ and let $U$ be an open set containing $( x_\alpha)$.

U comes from an open set in the regular product topology, so we can assume that it is $\prod U_\alpha \cap \prod'G_\alpha$ where only for finite number $U_\alpha \neq G$. the assumption that $I$ is infinite gives at least one $U_\beta=G$ so for all $(y_\alpha)$ such that $y_\alpha = 0 \forall \alpha\neq \beta$ we have $(x_\alpha)+(y_\alpha) \in U$ so $\sum (U) = G$, because we can choose $y_\beta$ to be any element of G.

This means that if $V\neq G$ is an open neighborhood of $\sum (x_\alpha)$ then there is no open neighborhood U of $(x_\alpha)$ such that $\sum (U) \subseteq V$.

Answer 2

It might be added that the example shows that the true coproduct of a non-finite number of copies of $\mathbb R$ does not have the subspace topology from the corresponding product. Indeed, on the usual categorical grounds, there is only (at most) one topology that fulfills the requirement (with regard to all possible mappings from the coproduct to all other topological groups/vectorspaces. As the example shows, the "correct" coproduct topology is considerably finer than the restriction of the product topology: for [Edit:] locally convex topological vector spaces it is essentially a "diamond" topology.

[Edit 2:] The diamond topology on a coproduct/sum of $V_i$ has local basis at 0 given by convex hulls of (image of) opens at $0$ in $V_i$. In the locally convex category, it is pretty clear that this has the desired property, thus constructing the coproduct.

[Edit: corrections related to "locally convex" modifier and "uncountable" modifier...]

The situation is not as trivial/boring as one might imagine, since uncountable coproducts of copies of $\mathbb R$ in the category of not-necessarily locally convex topological vector spaces are themselves not locally convex, due to the existence of not-locally-convex topological vector spaces, the $\ell^p(I)$ spaces with $0<p<1$ and possibly uncountable index set $I$. To see this, for a nbd $N$ of $0$ in the locally convex coproduct of copies of $\mathbb R$ indexed by uncountable $I$, with the diamond topology, for each $i\in I$ there is $\delta_i>0$ such that $N\cap \mathbb R_i\supset (-\delta_i,\delta_i)$. For $I$ uncountable, there is some $n_o$ such that there are infinitely-many $i_1,i_2,\ldots \in I$ with $\delta_i\ge 1/n_o$. Then the $p$-norms of the ever-larger convex combinations of $i_j$th-inclusion of $\delta_{i_j}$ are $\delta_{i_1}^p/n^p+\ldots+\delta^p_{i_n}/n^p$. These are bounded below by $n/n_o^pn^p=n^{1-p}/n_o^p$ which go to $+\infty$. This contradicts any hope for a continuous induced map to $\ell^p(I)$ with $0<p<1$.

This has the consequence that inductive limits of locally convex tvs's in the larger category of not-necessarily-locally-convex tvs's are not locally convex. Conceivably this sounds reasonable or innocent enough, but it entails that we can no longer appeal to Hahn-Banach, etc. At the very least, it certainly means that uncountable coproducts/colimits in the locally convex tvs category are not coproducts/colimits in the larger category.

This little story does also provide yet-another reminder of some of the unforeseen, non-elementary risks of infinitary operations, especially uncountable, I think.

Thx to Theo B. for correcting my earlier wild over-statement! :)