Re: Compact closure of half a subgroup

In article <Pine.BSI.4.58.0511190230170.24258@xxxxxxxxxxxxxxxxx>,
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:
>
>Let G be a topological group, g in G, A = { g^j | j = 0,1,.. }
>Assume cl A is compact. How to show cl A is a group?
>
>First off, notice (cl A)(cl A) subset cl AA = cl A.
>So cl A is closed under group operation.
>
>If perchance g^-1 in cl A, then from above
> for all k in N, g^-k in cl A.
>
>Thus H = < A > subset cl A.
> A subset H subset cl A; cl A = cl H.
>Finally
> (cl A)^-1 = (cl H)^-1 = cl H^-1 = cl H = cl A
>showing cl A closed under inverses.
>
>Thus for want of g^-1 in cl A, problem is sloved.
>Has anybody suggestions how to show g^-1 in cl A?

Suppose g^(-1) is not in cl A. Then there is a neighbourhood
U of the identity e such that g^(-1) U is disjoint from cl A.
Now for any nonnegative integer j, g^j U is disjoint from
g^(j+1) (cl A) = cl {g^k: k >= j+1} = cl A \ {g^k: 0 <= k <= j}.
But then if c is in cl(A) \ A, c U^(-1) is a neighbourhood of c
that contains no g^j. But this contradicts the definition of
closure. So cl(A) \ A is empty, i.e. A itself is compact.
But then the above neighbourhoods g^j U show that A is discrete
and therefore finite, but then it's easy...

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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