Ok, this is a really easy question, so I apologise in advance.(adsbygoogle = window.adsbygoogle || []).push({});

Let A be an abelian subgroup of a topological group. I want to show that cl(A) is also.

Now I've shown that cl(A) is a subgroup, that is fairly easy. So I just need to show it is abelian.

For a metric space, it is easy, since we can just write for any two elements h_1 h_2 in cl(H) as

h_1=lim_{i -> infty} (h_1(i))

h_2=lim_{i -> infty} (h_2(i))

where h_1(i) and h_2(i) are sequences converging too h_1 and h_2 respectively which are in H. Then to find h_1.h_2 we can just multiply the elements of these limits since multiplication is continuous and then we can commute them to deduce that h_1.h_2=h_2.h_1

Can I do something similiar for a general topological group, which is not a metric space? h_1 and h_2 are limit points, if they are not then they are in H anyway and will commute.

Maybe I just need my space to be Hausdorff, so then limits are unique?

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# Closure of an abelian subgroup

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