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It's sufficient show that for every point $x \in$ AB there exists a net $(x_\alpha)_{\alpha \in I}$ of points of AB such that $x_\alpha \rightarrow x$. Let $x = ab \in$ AB.

How A is closed exists a net $(a_\alpha)_{\alpha \in I}$ of points of A such that $a_\alpha \rightarrow a$. Analogously, how B is compact exists a net $(b_\sigma)_{\sigma \in J}$ of points of B such that some subnet $(b_{g(\beta)})_{\beta \in K}$ is such that $b_{g(\beta)} \rightarrow$ b.

Consider then the set $I \times K$. This is parcially ordered in the following way: $(\alpha, \beta) \le (\theta, \psi)$ iff $\alpha \le \theta$ and $\beta \le \psi$.

$\rho : I \times K \rightarrow G \times G$ given by $\rho(\alpha, \beta) = (a_{\alpha}, b_{g(\beta)})$. This application is a net that converges to $(a, b)$. Let $\cdot$ the operation of the group. Then $\cdot \circ \rho : I \times K \rightarrow G$ is a net of points of AB such that $\cdot \circ \rho \rightarrow $ ab.

Therefore, AB is closed. (Correct ?).

joriki
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Gleberson Antunes
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    Your proof is wrong from the very beginning. The condition for being closed is not that one (which is trivially satisfied for any subset, by taking a constant net). – Anne Bauval Jan 08 '23 at 23:58
  • @Anne Bauval, Why would it be wrong from the start? We know that given a point x and a set A, we have that x belongs to cl(A) iff there is a net of points from A that converges to x. – Gleberson Antunes Jan 09 '23 at 01:36
  • Yes, hence $A$ is closed iff for every converging "net of points from $A$", the limit belongs to $A.$ And not iff "for every $x∈ A$ there exists a net of points of $A$ converging to $x$" as you wrote: the latter condition trivially holds for every $A\subset G.$ – Anne Bauval Jan 09 '23 at 07:07
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    Yes, you are correct. Thanks a lot for correcting me. – Gleberson Antunes Jan 09 '23 at 12:20

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