
Topologising a Group
Posted:
Nov 21, 2012 11:28 PM


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?

