
Re: Topologising a Group
Posted:
Nov 22, 2012 7:06 AM


In article <Pine.NEB.4.64.1211212025320.22658@panix1.panix.com>, 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.
> > 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? >
 G. A. Edgar http://www.math.ohiostate.edu/~edgar/

