# Real-Analysis Real -numbers

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

a) If $0$ does not belong to $G$, then show that $H= \{xy|x,y ~\text{belong to}~ G\}$ is an open subset of $\mathbb R$.

b) If $0$ belongs to $G$, and if $(x+y)$ belongs to $G$ for all $x, y$ belonging to $G$ then show that $G=\mathbb R$.

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could not approach properly . plz help me out –  user61062 Feb 5 '13 at 19:49
Is R = $\mathbb{R}$, the set of all real numbers?, and is O = $0$ (zero?) –  amWhy Feb 5 '13 at 19:52
I'm assuming the letter O was used to mean $0$ zero. But please clarify if $R$ is intended to represent the set of real numbers: $\mathbb{R}$. I'm assuming that's the case, and will edit accordingly, but please do correct me if I'm wrong. –  amWhy Feb 5 '13 at 19:54
R- real numbers O- real number zero –  user61062 Feb 7 '13 at 21:17

Hint for a), choose $y \in G$ and let $\phi_y(x) = xy$. Note that $\phi^{-1}(z) = \frac{z}{y}$, hence $\phi_y$ is a homeomorphism.

Hence $\phi_y(G)$ is open, and hence $\cup_{y \in G} \phi_y(G)$ is open.

Hint for b), note that $(-\epsilon,\epsilon) \subset G$ for some small $\epsilon>0$, and $G+G \subset G$.

Hence we have $(-2\epsilon,2\epsilon) \subset G$, etc. Hence $\mathbb{R} \subset G$.

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For (b), use that $G$ is open to show that there must be an $\varepsilon>0$ so that the open interval $(-\varepsilon,\varepsilon)$ is contained in $G$, then use induction to show that for any natural number $n$, the interval $(-n\varepsilon,n\varepsilon)$ is contained in $G$.

Then show that for any real number $x$, there must be a natural number $n$ so that $|x|<n\varepsilon$.

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