Infinite products of topological groups - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T05:53:38Zhttp://mathoverflow.net/feeds/question/15199http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/15199/infinite-products-of-topological-groupsInfinite products of topological groupscommonname2010-02-13T18:03:12Z2010-02-14T01:47:01Z
<p>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:</p>
<p>$\pi_i \circ p = p_i$ for all $i \in I$.</p>
<p>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.</p>
<p>Is there any way to get around this?</p>
<p>Thanks.</p>
http://mathoverflow.net/questions/15199/infinite-products-of-topological-groups/15201#15201Answer by S1 for Infinite products of topological groupsS12010-02-13T18:20:24Z2010-02-13T23:12:57Z<p>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).</p>
<p>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 <a href="http://en.wikipedia.org/wiki/Product%5Ftopology" rel="nofollow">wiki</a> for the product topology).</p>