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What are the open sets in a topology specified by a fundamental system of neighbourhoods of $0$ of a topolgical group? Also, how is this topology unique. I searched this online, but the books I found only mention these statements, without specifying what the open sets are and why the topology is unique. Any help will be appreciated.

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Basically, because to define a topology we need to define a local base at every point in a consistent way. In a topological group $G$, for every $x$ and $y$ we have a homeomorphism of $G$ that maps $x$ to $y$, just use $h(z) = y*x^{-1}*z$. So we can transport a neighbourhood base of $e$ (the identity of $G$) to any other point of $G$, using such $h$. One then checks this is a consistent assignment and so determines the topology on $G$.

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Thanks. My issue is I don't understand what the open sets containing 0 are. Once I know these, I do understand I can get any open by translation and every open set arises this way. – user7288 Feb 20 '11 at 18:47
It all depends on how you define the topological group. Suppose you start with a group $G$ and then you need to define a topology; then it suffices to specify what the open neighbourhoods of $e$ are, or a base for those. Of course, one cannot do this arbitrarily, but there are theorems (in the standard textbooks) that tell you when such an assignment of potential neighbourhoods does give a topological group. I think most groups in practice are specified this way. The topology on the group need not be unique. But once a base is specified, it is uniquely determined. Can you give a concrete ex.? – Henno Brandsma Feb 21 '11 at 18:10

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