Neighborhoods in the product topology. In the last two lines thread, it is said that $N \times M \subseteq A\times B$ is a neighborhood of $0$ if and only if $N, M$ are neighborhoods of $0$. Here $A, B$ are topological abelian groups. How to prove this result? I searched on the Internet, but was not able to find a proof.
 A: If $A,B$ are topological spaces,then we define the product topology by means of the topologies of $A$ and $B$ in such a way that $U\times V$ is open whenever $U\subseteq A$ and $V\subseteq B$ are open. More proecisely, we take the smallest topology on $A\times B$ such that these $U\times V$ are all open. This is precisely achieved by declaring precisely the sets of the form
$$ \bigcup_{i\in I}U_i\times V_i$$
with $I$ an arbitrary index set and $U_i\subseteq A$, $V_i\subseteq B$ open.
If $C\subseteq A$, $D\subseteq B$ is such that $C\times D$ is open and nonempty(!), then we can conlude that $C,D$ are open. Why?
Assume 
$$ C\times D = \bigcup_{i\in I}U_i\times V_i$$
as above.
Since $C\times D$ is nonempty, let $(c,d)\in C\times D$. From the way $\{c\}\times D\subseteq C\times D$ is covered by the $U_i\times V_i$, we observe that
$$ D=\bigcup_{i\in I\atop c\in U_i}V_i $$
hence $D$ is open. Of course $d\in D$ so that $D$ is an open neighbourhood of $d$.
Similarly, $C$ is an open neighbourhood of $c$.
A: It is important to note that in that thread it is explicitly said that $\,G\times G\,$ has the product topology, meaning that a subset here is open iff it is the union of sets of the form $\;N\times M\;$ , with $\;N,M\subset G\;$ open...This, I think, solves the problem, as it is exactly the same with $\,A\times B\;$ a product of top. groups  :)
