Vtyjkyuk

If $G$ is an open set of $mathbb R$ which does not contain $0$, then how to prove that the set $G^2 = xy : x , y in G$ is also open?

And why $0$ has to be excluded?

Can anyone please help me by giving some hints.

If $p neq 0$ is a real number, $m_p:mathbbR to mathbbR$ defined by $m_p(x) = px$ is a homeomorphism with continuous inverse $m_frac1p$. In particular all such $m_p$ are open maps.

Then note that $$xy: x, y in G = bigcup_ p in G m_p[G]$$
which is a union of open sets, hence open.

$m_0$ is the constantly $0$ map, so not an open map, so this proof does not work if $0 in G$. I don't know off hand a simple counterexample that this would fail for just any open $G$ in the reals.

Actually, you don't have to exclude $0$, it just makes the proof a bit easier.

If $G$ is an open set that does contain $0$, $G^2 = (G backslash 0)^2 cup 0$. We already know $(G backslash 0)^2$ is open. Now for some $epsilon > 0$, $G$ contains $(-epsilon, epsilon)$, so
$G^2$ contains $(-epsilon^2, epsilon^2)$ and thus a neighbourhood of $0$.