Corollary. Let \(G\) be a topological group; let \(A\) and \(B\) be subsets of \(G\). If \(A\) is closed in \(G\) and \(B\) is compact, then \(A \cdot B\) is closed in \(G\). [Hint: First give a proof using sequences, assuming that \(G\) is metrizable.]
Solution
Let \(x\) lie in the closure of \(A\cdot B\). By the closure-net theorem, choose a net \[ a_\alpha b_\alpha\to x \] with \(a_\alpha\in A\) and \(b_\alpha\in B\). Since \(B\) is compact, the net \((b_\alpha)\) has an accumulation point \(b\in B\); equivalently, it has a subnet \(b_{\alpha_\beta}\to b\). Along the same subnet, \[ a_{\alpha_\beta}b_{\alpha_\beta}\to x. \] Continuity of multiplication and inversion gives \[ a_{\alpha_\beta}=(a_{\alpha_\beta}b_{\alpha_\beta})b_{\alpha_\beta}^{-1}\longrightarrow xb^{-1}. \] Since \(A\) is closed and all \(a_{\alpha_\beta}\in A\), we have \(xb^{-1}\in A\). Therefore \[ x=(xb^{-1})b\in A\cdot B. \] Thus \(A\cdot B\) is closed.