```Date: Nov 21, 2012 11:28 PM
Author: William Elliot
Subject: Topologising a Group

Let F be a filter over a group G, with e in /\F.For all g in G, let B_g = { gU | g in G }.Notice that B_g can be taken as a base for gand B = \/{ B_g | g in G } as a base for a topology for G.Accordingly give G the topology tau, generated by B.The inverse function, i:G -> G, g -> g^(-1) is continuous.Proof.  Assume g^(-1) in open U.  Thus g in open U^(-1) = { h^-1 | h in U }Since i(U^-1) = U, we see that i is continuous at g.Subsequently, i is continuous.By construction if U is an open set, then for all g in G, gU is an open set.Lastly, for (G,tau) to be a topological, the group operation p:G^2 -> G, (a,b) -> ab needs to be continuous.  Is the current premise sufficient?I think not.  Accordingly assume G is Abelian and that for all U in F, there's some V in F with VV = { uv | u,v in V } subset U.With those assumptions, the group operation p, is continuous.Proof.   Assume a,b in G and ab in open U.  Thus e in b^-1 a^-1 U and there'ssome open V nhood e with VV subset b^-1 a^-1 U.Subsequently a in open aV, b in open bV andp(aV,bV) = (aV)(bV) = abVV subset abb^-1 a^-1 U = U.Thusly for all a,b in G, p is continuous at (a,b) and p continuous.Can the assumptions be weakened or made simpler?
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