Proposition 9.1.8.label Let $G$ be a topological group and $A, B \subset G$, then
- (1)
If $A$ is open, then $AB$ is open.
- (2)
If $A$ is closed and $B$ is compact, then $AB$ is closed.
Proof, [I.1.1, SW99]. (1): For every $x \in B$, $Ab$ is open by translation invariance, so
is open.
(2): Let $x \in \overline{AB}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap AB \ne \emptyset$, so $UB^{-1}\cap A \ne \emptyset$, and $\fB = \bracsn{UB^{-1}| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
By Proposition 9.1.7, $\overline{UB^{-1}}\subset UUB^{-1}$, so
Since $\fF$ converges to $x$, (TG1) implies that $\bracs{UU| U \in \fF}$ contains a neighbourhood base of $x$. Thus
so $x \in yB \subset AB$.$\square$
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