# Infinite products of topological groups

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?

Thanks.

• The tag [topological-groups] already exists. I edited your question for you though. Feb 13, 2010 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. Feb 13, 2010 at 23:12
• "...nonempty family of NONEMPTY topological spaces" :) Feb 14, 2010 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. Jan 22, 2012 at 15:31

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).