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 g
and 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's
some open V nhood e with VV subset b^-1 a^-1 U.

Subsequently a in open aV, b in open bV and
p(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?