# A product of a compact subset and a closed subset is closed

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 $(0,0) \notin A + B$ .

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