If $G$ be an open set, then show that $H=\{xy: x, y \in G \}$ is open in $\mathbb{R}$

Question

Let $G$ be an open subset of $\mathbb{R}$.

1. If $0 \notin G$ then show that $H=\{xy: x, y \in G \}$ is an open subset in $\mathbb{R}$.

2. If $0 \in G$ and if $x+y \in G$, for every $x, y \in G$ then show that $G = \mathbb{R}$.

Attempt:

1. $G$ being open, $\forall x \in G$, $\ \exists \epsilon>0$, such that $(x-\epsilon, x+ \epsilon) \subset G$.

Let $T = \{zy: z\in (x- \epsilon, x+\epsilon), y \in \mathbb{R} \setminus \{0\} \}$. $T$ is clearly a connected and an open set, since when $y>0$, $ \ x- \epsilon<z < x+\epsilon \implies xy-y \epsilon<zy<xy+y \epsilon$, similarly the case in $y<0$.

Now, $H= \displaystyle[\bigcup_{0<y \in G} {(xy-y \epsilon,xy+y \epsilon)}] \displaystyle\bigcup[\bigcup_{0>y \in G} {(xy+y \epsilon,xy-y \epsilon)}]$

Clearly, $H$ is open, being an infinite union of open sets.

2. Since $0$ is in $G$ and the set is open, $\exists$ some $\epsilon>0$ such that $(-\epsilon, \epsilon) \subset G$.

Now, $ -\epsilon< s, r < \epsilon \implies -2 \epsilon < s+r < 2 \epsilon.$. By the Archimedean property, any $N \in \mathbb{R}$ can be made less than $n \epsilon$ by suitable choice of $n \in \mathbb{N}$, i.e. $(-N,N) \subset (-n \epsilon, n \epsilon)$ $\forall N \in \mathbb{R^+}$. Hence, $G = \mathbb{R}$.

Please verify.

Answer 1

If $y \neq 0$ then $\{yx \}_{x \in G}$ is open. Hence $\cup_{y \in G, y \neq 0} \{yx \}_{x \in G}$ is open.

If $0 \notin G$, then it is clear that $\{xy\}_{x,y \in G}$ is open.

If $0 \in G$, then $(-\epsilon, \epsilon) \subset G$ for some $\epsilon>0$ and so $0 \in {1 \over 2} \epsilon \cdot (-\epsilon, \epsilon) \subset {1 \over 2} \epsilon \cdot G$, and so $0 \in \cup_{y \in G, y \neq 0} \{yx \}_{x \in G}$ and so $\cup_{y \in G} \{yx \}_{x \in G} = \cup_{y \in G, y \neq 0} \{yx \}_{x \in G}$ and hence it is open.

In the latter case, since $n G \subset G+\cdots + G = G$, we see that $(-n\epsilon, n \epsilon) \subset G$ for all $n$.