# Closure of an abelian subgroup

1. Dec 13, 2009

### Jamma

Ok, this is a really easy question, so I apologise in advance.

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?

2. Dec 13, 2009

### Jamma

Oh, sorry, its going to be second (which implies first) countability that I will need. That will allow me to write the points as limits. Sorry!

3. Dec 17, 2009

### g_edgar

Yes, you can use convergence of nets (or filters) to do the proof. But, as you note, Hausdorff is required. Indeed, the result is, in general, false in non-Hausdorff topological groups.