Let $G$ be an open subset of $\mathbb{R}$ . Then:
a) Is the set $H=\{xy|x,y\in G\ \text{and}\ 0\notin G\}$ open in $\mathbb{R}$?
b) Is the set $G=\mathbb{R}$ if $0\in G$ and $\forall x,y\in G, x+y\in G$?
I think the answer to both the problems is yes. But, the question is, should we use the group theoretic properties or topological properties to prove the statements? And how should we exactly proceed. Any hints? Thanks beforehand.
On
We're asking if a set is open, so we're going to have to use at least some topological properties. For a, I'd use the fact that multiplication by a fixed $x\neq0$ is a homeomorphism. So then our set is a union (over $x\in G$) of sets homeomorphic to $G$.
b I'd approach differently. Here we can use the Archimedean property. In particular, $G$ must contain some interval $[-\varepsilon,\varepsilon]$. Every real number is a finite positive integer multiple of something in that interval.
What "group properties" could possibly be of any use? Especially as no-one claimed $G$ was a sub group of $\mathbb R$.
1) I'd use that that standard that because $G$ is open, then for any $x \in G$ there is an $\epsilon_x>0$ so that for any $x': x-\epsilon_x < x' < x + \epsilon_x$ we know $x'\in G$ and for any $y \in G$ there is an $\epsilon_y>0$ so that for any $y': y-\epsilon_y < y' < y + \epsilon_y$ we know $y'\in G$.
Does it follow that for $xy\in H$ that there is a $\epsilon_{xy}> 0$ so that for any $z: xy-\epsilon_{xy} < z < xy + \epsilon_{xy}$ that $z \in H$?
That should be straightforward (albeit tedious) to prove.
2) is different. Hint: If $x \in G$ then by induction $n*x \in G$ for all $n\in\mathbb N$. Use archimedian property and the fact that for all $\epsilon > 0$ there is an $0 < x < \epsilon; x \in G$.