A question on topological group

Question

Let $G$ be a topological group. We know that if $A$ is a closed subset of $G$ and $B$ a compact subset of $G$, then $A+B$ is a closed subset of $G$. My question: Is the above statement true whenever $B$ is a $\sigma-$compact set of $G$? Thank you

Answer 1

You asked in your comment here whether the claim is true if $B$ is additionally closed, i.e. you want a counterexample to:
$A$ is closed, $B$ is $\sigma$-compact and closed $\Rightarrow$ $A+B$ is closed.

Take, for example: $$A=B=\{(x,y)\in\mathbb R^2; y\ge1/|x|\}$$ in $(\mathbb R^2,+)$ (with the usual - coordinatewise - addition). Then $A+B=\{(x,y)\in\mathbb R^2; y>0\}$.

Answer 2

No. Take $G$ to be the unit circle in the complex plane, $A=\{1\}$ and $$B=\{e^{n\pi i\theta}\mid n\in\mathbb Z\}$$ where $\theta$ is an irrational number. $B$ is $\sigma$-compact because it is countable. $B$ is not closed, in fact it is dense.

Answer 3

Not in general: take $G=\Bbb R$, $A=\{0\}$, and $B=(0,1)$.