The product $\prod_jG_j$ of the <finite groups> with their <discrete topology> is a <topological group> under coordinatewise multiplication and inversion. The compatibility equations defining $G=\varprojlim_jG_j$ are preserved by both operations, so $G$ is a subgroup. Their restrictions to the <subspace topology> on $G$ are continuous. Hence $G$ with its standard <inverse-limit topology> is a topological group.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 1 / b / i / Solution

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For $A\subseteq X$, the subspace topology consists of the intersections $A\cap U$ with open subsets $U\subseteq X$.


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 Incoming links: Subspace topology