Subsets of Topological Hausdorff Groups

Question

Let $A, B$ be two subsets of a topological Haudorff Group. Need to show that

• $\forall B: A$ open $\Rightarrow$ $AB$ open
• $A,B $ compact $\Rightarrow$ $AB$ compact
• $A$ closed and $B$ compact $\Rightarrow AB$ closed
• $A,B $ closed than $AB$ does not necesarry need to be closed

So far i know that every $T1$ Group is a Hausdorff (right?). I dont know how to do this since I have not subgroups here but subsets... Anyone an idea?

Thanks

Answer 1

Let $G$ be your group.

• $AB=\bigcup_{b\in B}Ab$, which is an union of open sets.
• $AB$ is the image of $\pi\colon A\times B\longrightarrow G$ defined by $\pi(x,y)=xy$ and $A\times B$ is compact.
• There is an answer here.
• Take, in $(\mathbb{R},+)$ (endowed with its usual topology) $A=\mathbb{N}$ and $B=\left\{-n+\frac1n\,\middle|\,n\in\mathbb{N}\right\}$. Then $A$ and $B$ are closed, but $A+B$ is not (it contains the sequence $\left(\frac1n\right)_{n\in\mathbb N}$, but not its limit).