I came across the following statement while reading notes in a course about p-adic representations:

Let $ G$ be a topological group and let $ A \subseteq G $ be a compact subset and $ B \subseteq G$ be a closed subset. Then $ A \cdot B \subseteq G$ is closed.

Although the claim doesn’t seem to be too difficult, I didn’t find the proof to be immediate. I tried googling a bit and found the following math.stackexchange question, but the accepted answer there is wrong. Furthermore, the solution assumes that the space is regular, which was fine in my case since I was dealing with a metrizable space, but the statement is true also without this assumption. I’ve found this solution which I want to explain.

We want to show that $ G \setminus \left({A \cdot B}\right) $ is open. Let $ x \in G \setminus \left({A \cdot B}\right) $ . For every $ a \in A $ we have that $ a^{-1} \cdot x \notin B $ . Consider the map $ m:G \times G \rightarrow G $ defined by $ m(g,h) = g^{-1} \cdot h $ . Then $ m$ is continuous. Since $ B $ is closed, $ G \setminus B $ is open, and therefore $ m^{-1}(G \setminus B) $ is open. Since for every $ a \in A $ , we have $ m(a,x) = a^{-1} \cdot x \notin B $ , we have $ (a, x) \in m^{-1}(G \setminus B)$ and since the latter is open, there exists open subsets of $ G $ with $ a \in U_a, x \in V_a $ such that $ m(U_a \times V_a) \subseteq G \setminus B $ .

Consider the following open cover of $ A $ : $ A \subseteq \bigcup_{a \in A} U_a $ . Since $ A $ is compact, this cover admits a finite sub-cover $ A \subseteq \bigcup_{i \in 1}^N U_{a_i} $ . Define $ V = \bigcap_{i = 1}^N V_{a_i} $ . Then $ V $ is open with $ x \in V $ . We claim that $ V \subseteq G \setminus \left({A \cdot B}\right) $ : otherwise there exists $ a \in A, b \in B $ with $ a \cdot b \in V $ . Since $ a \in A $ , we have that $ a \in U_{a_i} $ for some $ i $ . Then $ ab \in V \subseteq V_{a_i} $ and $ m(a,ab) = a^{-1} \cdot (ab) = b \in B $ , which is a contradiction since $ U_{a_i} \times V_{a_i} \subseteq m^{-1}(G \setminus B)$ .

Note that this proof doesn’t assume that $ G $ is Hausdorff.

Also note that the claim isn’t true if the assumption that $ A $ is compact is replaced with the assumption that $ A $ is closed: we can take in this case $ A = \{ (t, \frac{1}{t} ) \mid t > 0 \} $ , $ B = \{ (t, -\frac{1}{t} ) \mid t > 0 \} $ . The set $ A + B$ clearly contains the sequence $ (\frac{1}{n}, 0) $ but $ (0,0) \notin A + B $ .