Let $\delta = (\delta_1,\ldots,\delta_d)$ be a fixed $d$-tuple with $\delta_i > 0$ for $i=1,\ldots,d$. For a set $E \subset \mathbb{R}^d$, define
$$\delta E := \{(\delta_1 x_1,\ldots,\delta_d x_d) : (x_1,\ldots,x_d) \in E \}.$$
Show that if $E$ is open then $\delta E$ is open.
My attempt: (Here, I use the notation $\delta x$ to denote the point $(\delta_1 x_1,\ldots,\delta_d x_d)$ for each $x = (x_1,\ldots,x_d)$.) Let $x \in \delta E$. We want to show that there exists some $r' > 0$ such that $B_{r'}(x) \subset \delta E$. Since $x \in \delta E$ we have $x = \delta y$ for some $y \in E$. Since $E$ is open, there exists some $r > 0$ such that $B_r(y) \subset E$. Here, I'm a bit stuck. I've considered $r' = r |\delta|$, but it doesn't quite seem to work. If $z \in B_{r |\delta|}(x)$ then $|z – x| = |z – \delta y| < r |\delta|$, which implies $|\frac{z}{|\delta|} – \frac{\delta}{|\delta|} y |< r$. Then I was hoping to use the fact that $E$ is open here, but I am unfortunately stuck. Any help would be appreciated.
Note: I think the result is immediate from the fact that $(x_1,\ldots,x_d) \mapsto (\delta_1 x_1,\ldots,\delta_d x_d)$ is a continuous map, but I was just wondering if there's a way to prove this using open balls.
Best Answer
See this. Cartesian product of homeomorphisms is always a homeomorphism in product space.
Then to prove your proposition, we only need to verify that the following function is a homeomorphism. $f_i:\mathbb{R}\to\mathbb{R}, f_i(x)=\delta_ix$.
In Euclidean space, polynomial funciton is always a continuous function, and $f_i^{-1}$ is also polynomial function.
Then $f_i$ is homeomorphism.
$E=(\prod_1^df_i)^{-1}\delta E$, so if $\delta E$ is open, then $E$ is also open.