# Properties of open subsets of the real numbers

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.

• It's a topology question so use topological properties. Namely if $x,y\in G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product? Nov 19, 2018 at 18:30
• Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $\mathbb R$ because $G$ is not a group) could be useful. Nov 19, 2018 at 18:32

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 $$0 < x < \epsilon; x \in G$$.

• thanks! great answer without using homeomorphisms for the first part Nov 22, 2018 at 7:10

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.

• edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here? Nov 19, 2018 at 17:52
• @vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck. Nov 19, 2018 at 17:57