If $G\subset\mathbb R\setminus\{0\}$ is open, then $G^2= \{xy : x\in G , y \in G\}$ is open

Question

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.

Answer 1

If $p \neq 0$ is a real number, $m_p:\mathbb{R} \to \mathbb{R}$ 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.

Answer 2

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 $0$.