Let $G$ be a topological group and $(x_j)_{j \in J}$ and $(y_j)_{j \in J}$ be two nets in $G$ indexed by the same directed set $J$ converging to $x$ and $y$ in $G$ respectively. Can it be said that $x_jy_j \to xy\ $?
$\textbf {My Attempt} :$ I am trying approach in the following way. Let $N$ be a neighbourhood of $xy$ in $G.$ Then $x^{-1} N$ is a neighbourhood of $y$ and $Ny^{-1}$ is a neighbourhood of $x.$ Since $x_j \to x$ there exists $j_0 \in J$ such that for all $j \geq j_0$ we have $x_j \in N y^{-1}.$ Also since $y_j \to y$ there exists $j_0' \in J$ such that for all $j \geq j_0'$ we have $y_j \in x^{-1} N.$ Now since $J$ is a directed set there would exist a $j' \in J$ such that $j' \geq j_0$ and $j' \geq j_0'.$ Now by transitivity of ''$\geq$'' it follows that for all $j \geq j'$ we have $x_j \in N y^{-1}$ and $y_j \in x^{-1} N$ and hence $x_j y_j \in N y^{-1} x^{-1} N = N (xy)^{-1} N.$ Now I got stuck and could not proceed further. How do I show from here that the net $(x_jy_j)_{j \in J}$ is eventually in $N\ $?
A hint is very much required at this stage. Thanks!
$\textbf {EDIT} :$ I can show that $N \subseteq N(xy)^{-1}N.$ Because for any $z \in N$ we can write $$z = z\ (xy)^{-1} (xy) \in N (xy)^{-1} N.$$ But how do I prove the converse?
You should start the proof with continuity of the map $(x,y) \to xy$. Given a neighborhood $N$ of $xy$ there exist neighborhoods $N_1$ and $N_2$ of $x$ and $y$ respectively such that $N_1 . N_2 \subseteq N$. I believe the rest of the proof easy for you now.