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While studying for a topological groups course, I wondered if we could define the product of uncountably many topological groups such that the product is still a topological group. That is: let $G_i$ be a topological group with product law $p_i$ for each $i \in I$ (with $I$ uncountable). We can give $G = \prod_{i \in I} G_i$ the (Tychonoff) product topology and define the product law of $G$ by:

$\pi_i \circ p = p_i$ for all $i \in I$.

However, when trying to prove that this mapping is continuous end up needing $I$ to be at most countable or that the topologies of $G_i$ be discrete.

Is there any way to get around this?


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The tag [topological-groups] already exists. I edited your question for you though. – Harry Gindi Feb 13 '10 at 18:11
It is true though that some nice properties of topological spaces are lost by taking uncountable products. For instance, given a nonempty family of topological spaces, uncountably many of which are not endowed with the trivial topology, the product is not first-countable. In particular, metrizability is lost in uncountable products. – Pete L. Clark Feb 13 '10 at 23:12
"...nonempty family of NONEMPTY topological spaces" :) – Pete L. Clark Feb 14 '10 at 1:48
... not that $G_i$ are compact implies that the product is compact. It is often an issue that this is not true for locally compact. here, one prefers the restricted product. – Marc Palm Jan 22 '12 at 15:31
up vote 8 down vote accepted

You can define the product of an arbitrary family $(G_i)_{i \in I}$ of topological groups $G_i$ by equipping the group-theoretic product $G = \prod_{i \in I} G_i$ with the product topology; the product topology is indeed compatible with the group structure (confer Bourbaki, General topology, III.2.9, but it's pretty obvious actually).

Perhaps your problem is the product topology? Note that a basis for the product topology are the sets $(U_i)_{i \in I}$ where $U_i \subseteq G_i$ is open and $U_i = G_i$ for all but finitely many $i \in I$. (confer wiki for the product topology).

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...where, in this context, "almost all" means "all but finitely many". – Gerald Edgar Feb 13 '10 at 18:29
I took the liberty of editing as GE suggested above, an instance of my "easy question deserves easily read answer" philosophy. @Arminius, if for any reason you truly prefer your phrasing, please feel free to change it back; you needn't defend your reasons for doing so. – Pete L. Clark Feb 13 '10 at 23:15
No, no, it's fine. But just for my own education: Is there a difference between "almost all" and "all but finitely many"? For me it's (at the moment) exactly the same... – user717 Feb 13 '10 at 23:43
Arminius, I guess the point is that "almost all" to mean "all but finitely many" is a slightly informal usage. At least, that's been my own experience. So if someone didn't know what it meant, they might have trouble looking it up. Indeed, they'd probably find lots of stuff defining "almost all" in the measure theory sense. – Tom Leinster Feb 13 '10 at 23:59
@Arminius and TL: right, for many students "almost all" is most familiar in the sense of measure theory, so I can imagine someone looking for a suitable measure on the arbitrary index set I. (Of course "almost all" in this sense is equivalent to the set of exceptions having finite counting measure, but that seems overly convoluted...) – Pete L. Clark Feb 14 '10 at 1:17

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