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