# A question in topological group

• Sep 16th 2010, 12:54 AM
luoginator
A question in topological group
Hey guys, I was just thinking of a question in topological group. Suppose A and B are two closed sets in a topological group, then A*B is closed. Can any of you help me think of a counter example of this please? Or, if you think it is true, then you may prove it.

Thanks a lot.
• Sep 16th 2010, 04:19 AM
Opalg
Quote:

Originally Posted by luoginator
Hey guys, I was just thinking of a question in topological group. Suppose A and B are two closed sets in a topological group, then A*B is closed. Can any of you help me think of a counter example of this please? Or, if you think it is true, then you may prove it.

This need not be true even in an Abelian group. For example, a normed vector space is a topological group under addition (just ignore the scalar multiplication), and it is possible for the sum of two closed subspaces to be non-closed. A specific example of that is given here.
• Sep 16th 2010, 03:52 PM
luoginator
Thanks, but what about the multiplication. A*B means a*b for a belongs to A and b belongs to B. Thanks
• Sep 16th 2010, 07:13 PM
Iondor
I have a counterexample:
Take the multiplicative group of the reals
$G:=\mathbb{R}-\{0\}$

$
A:=\mathbb{Z}-\{0\}
$

$
B:=\{\frac{1}{n} : n \in \mathbb{N} , n\geq 1\}
$

are closed in G, but
$A\cdot B=\mathbb{Q}-\{0\}$
isn't.
• Sep 16th 2010, 09:40 PM
luoginator
Nice Counter Example. thx:)
• Sep 16th 2010, 11:55 PM
Opalg
Quote:

Originally Posted by luoginator
Thanks, but what about the multiplication. A*B means a*b for a belongs to A and b belongs to B. Thanks

A vector space is an additive group, so the group operation is addition.

Iondor's example is a lot simpler and neater than the one I gave. But the vector space example shows that you can even take A and B to be closed subgroups of a topological group, and their "product" (which is their sum in this case) can still fail to be closed.