
Re: Topologising a Group
Posted:
Nov 22, 2012 10:29 PM


On Thu, 22 Nov 2012, G. A. Edgar wrote: > William Elliot <marsh@panix.com> wrote: > > > 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. > > No good. Example. The group is the real numbers with addition. > The filter F consists of the sets [0,a) with a>0 . Then in your > topology the inverse function is not continuous. An assumeption is needed, namely for all U in F, U^1 in F. Thus the inverse would be continuous.
In addtion, in the proof below for continuity of the binary operator, the assumption of Abelain can be replace with the weaker for all g in G, U in G, gUg^1 in F. in other words, Ug is open.
Are those the simpliest premises for topologising a group?
> > 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?

