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