$D$ is a subset of $\mathbb R$ and $E=\{\alpha x : x\in D, \alpha>0\}$
Prove that $E$ is open iff $D$ is open
For each $\alpha x \in E \exists$ a ball $B_\epsilon (\alpha x)\subset E$. Can we conclude that there is a ball of radius $1/ \alpha$ at the corresponding $x\in D$?
Similarly for each $B_\epsilon(x)\subset D$, can we say $B_{\alpha \epsilon} (\alpha x) \subset E$?
Does this proof work?