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Topic: Topologising a Group
Replies: 6   Last Post: Nov 25, 2012 10:39 PM

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G. A. Edgar

Posts: 2,510
Registered: 12/8/04
Re: Topologising a Group
Posted: Nov 22, 2012 7:06 AM
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In article <>,
William Elliot <> 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

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