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.
Best Answer
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).