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?


  • 1
    $\begingroup$ The tag [topological-groups] already exists. I edited your question for you though. $\endgroup$ Feb 13, 2010 at 18:11
  • $\begingroup$ 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. $\endgroup$ Feb 13, 2010 at 23:12
  • $\begingroup$ "...nonempty family of NONEMPTY topological spaces" :) $\endgroup$ Feb 14, 2010 at 1:48
  • $\begingroup$ ... 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. $\endgroup$
    – Marc Palm
    Jan 22, 2012 at 15:31

1 Answer 1


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).

  • $\begingroup$ ...where, in this context, "almost all" means "all but finitely many". $\endgroup$ Feb 13, 2010 at 18:29
  • $\begingroup$ 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. $\endgroup$ Feb 13, 2010 at 23:15
  • $\begingroup$ 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... $\endgroup$
    – user717
    Feb 13, 2010 at 23:43
  • $\begingroup$ 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. $\endgroup$ Feb 13, 2010 at 23:59
  • 2
    $\begingroup$ @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...) $\endgroup$ Feb 14, 2010 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.