Let $A,B$ be subsets of a topological group $G$. If $A$ is closed and $B$ is compact then $A\cdot B$ is closed. (Topolgy, munkres, ch3 sec26 exer13)

Question

I solved this problem, but I'm not sure.

$Problem$

Let $G$ be a topological group and $A,B$ are subspaces of G. If $A$ is closed and $B$ is compact, then $A\cdot B$ is closed

$Sol$

Suppose $c\notin A\cdot B$ then for each $a\in A$ and $b\in B$, $c\neq ab$ so that $$cb^{-1}\neq a.$$ It follows that $$cB^{-1}\cap A=\emptyset.$$ Note that $cB^{-1}$ is compact and $A$ is closed. For any $b\in B$, since $cb^{-1}\notin A$ and G is regular, there are disj nbds $U_b,V_b$ of $cb^{-1}, A$, repectively. Cover $cB^{-1}$ by the open sets $U_b$ and choose a finite subcover, say $U_{b_1},...,U_{b_n}$. Let $U=\cup U_{b_i}, V=\cap V_{b_i}$ so that they are disj nbds of $cB^{-1}, A$, respectively. We assert that $U\cdot B,V\cdot B$ are disj nbds of $c, A\cdot B$, respectively. For, $$c=cb^{-1}b\in U\cdot B$$ and $$AB\subset V\cdot B$$ and $$U\cdot B \cap {V\cdot B}=(U\cap V)\cdot B=\emptyset\cdot B=\emptyset$$So $c$ is an exterior point of $A\cdot B$, it follows that $A\cdot B$ is closed.

I don't feel comfortable because I didn't use a hint, is it right that I solved it? And is there a solution using given 'hint'

"If c is not in $A\cdot B$, then there is a nbd $W$ of c such that $W\cdot B^{-1}$ is disjointed from $A$"

Answer 1

Let $f$ be the operation on $G\times G$, so $f$ is continuous. Suppose $c\notin AB$, then of course $cB^{-1}\cap A=\emptyset$. Note that $B^{-1}$ is compact and $f^{-1}(A)$ is closed in $G\times G$ and so the complement $X={f^{-1}(A)}^c$ is open. Since $(c,b^{-1})\in X$, we can choose, for each, $b\in B$ nbds $W_b,V_b$ of $c,b^{-1}$, respectively, such that $$(c,b^{-1})\in W_b\times V_b\subset X.$$Cover $B^{-1}$ by $V_b$, and so we have a finite cover, say $V_{b_1},...,V_{b_n}$. Let $$W=\cap W_{b_i}, V=\cup V_{b_i}.$$Now, if $x\in WB^{-1}$, then we can write$$x=wb^{-1}$$ where $w\in W$ and $b^{-1}\in B^{-1}.$ Then for some $i$, $w\in W_{b_i}$ and $b^{-1}\in V_{b_i}$, so $x\in A^c$. i.e., $WB^{-1}\cap A=\emptyset$

Answer 2

Simpler alternative if you know nets: let $(a_i \cdot b_i)_{i \in I}$ be a net in $A\cdot B$ that converges to $g \in G$. As $b_i$ is a net in the compact $B$ there is a convergent subnet $(b_j)_{j \in J}$ that converges to some $b \in B$. But then $a_j = (a_j\cdot b_j) \cdot b_j^{-1} \to g \cdot b^{-1}$ by continuity of group operations, and as $A$ is closed, it follows that $g \cdot b^{-1} \in A$ and so $g = (g \cdot b^{-1}) \cdot b \in A \cdot B$, as required. Hence the latter set is closed. I find this proof an illustration of the elegance of nets in proofs like this.