Vtyjkyuk

This exercise comes from Munkres' Topology, 2nd edition, Page 188. It says

11. 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 $Acdot B$ is closed in $G$.[ Hint: First give a proof using sequences, assuming that $G$ is metrizable. ]

This seems to be nontrivial to me. I know there is a proof without using nets, but I'm fascinated by the way with nets. I have been pondering this exercise all the whole day and cannot figure out how to use nets to prove it. Currently I learn topology on my own so I did some research, but it still resists my efforts to solve it. There is a hint but I cannot benefit from it. I need more hints, or even a proof using nets. Any help will be appreciated. Thanks.

A proof using nets is straightforward enough: suppose $f: I to G$ is a net from a directed set $(I,le_I)$ to $G$ such that all $f(i) in Acdot B$ and suppose that $f to p in G$. We need to show that $p in A cdot B$ as well (a set is closed iff it's closed under limits of nets).

We start by writing $f(i) = a_i cdot b_i$ for some $a_i in A, b_i in B$ for each $i in I$. Then $f_B: I to G$ defined by $f_B(i) = b_i$ is a net with values in the compact set $B$ so by a characterisation of compactness using nets, there is a subnet $g_B: J to G$ of $f_B$ and some $b_0 in B$ such that $g_B to b_0$, and where $(J, le_J)$ is some directed set and there is a connection function $c: J to I$ witnessing the subnet relation.

First we conclude that the net $j to f(c(j))g_B(j)^-1$ converges to $p cdot b_0^-1$
by continuity of the group operation. As all $f(c(j))g_B(j)^-1 in A$ we conclude by closedness of $A$ that $pcdot b_0^-1 in A$, so that $p = (p cdot b_0^-1)cdot b_0 in A cdot B$, as required.

Note that it uses the exercises 6,7 and 10 before it. (This is exercise 11). It's a nice exercise to show the naturalness of net-proofs, IMHO.

This is the proof for sequences: We want to show that the limit of any convergent sequence in $Acdot B$ is in $Acdot B$.

So if $a_nb_nin Acdot B$ and $a_nb_nto gin G$, then we want to show $gin Acdot B$. Choose a subsequence such that $b_n_kto bin B$ (using compactness). Then $a_n_k=a_n_kb_n_kb_n_k^-1to gb^-1$ so (as $A$ is closed) $gb^-1in A$. Hence $g=gb^-1bin Acdot B$. So $Acdot B$ is closed.

Now can you rewrite this with nets instead of sequences?