Every open ball in $\mathbb{R}^n$ is a union of balls with rational centers and rational radii.
My proof:
Let $B(x,r)$ be an open ball in $\mathbb{R}^n$. Let $y\in \mathbb{Q}^n$ be such that $|x_i-y_i|<\frac{r}{2n}$ for each $i=1,2,3…,n$(this is possible because the rationals are dense in the reals). Then, if $a\in B(y,\frac{r}{2\sqrt(n)})$, it follows that $d(x,a)\leq d(x,y)+d(y,a)$ $<$ $\frac{r}{2\sqrt{n}}$ $+$ $\frac{r}{2\sqrt{n}}$ $=$ $\frac{r}{\sqrt{n}}$ $\leq r$. Hence, $B(y,\frac{r}{2\sqrt(n)})$ $\subseteq B(x,r)$. Since for each $a\in B(x,r)$, there exists a rational radius $r_x>0$ such that $B(a,r_x)\subseteq B(x,r)$, $B(x,r)= \bigcup_{a \in B(x,r)}B(a,r_x)$. For each $B(a,r_x)$, choose a rational $y\in \mathbb{Q}^n$ such that $B(y,\frac{r_x}{2\sqrt(n)})$ $\subseteq B(a,r_x)$. This means that $B(a,r_x)$ is the union of all possible such balls. Hence $B(x,r)$ is the union of balls with rational centers and rational radii.
Is my proof correct?
Best Answer
1) Yes, your proof is correct.
2) Your proof is unnecessarily complicated/written in an unnatural way. I would say:
Let $B=B(x,r)$ be an open ball, and let $z\in B$. We are going to build a rational ball $B'$ such that $z\in B'\subset B$.
Since $z\in B$ and $B$ is open, there is an open ball centered at $z$ and contained in $B$: $B(z,r_z)\subset B$. We are free to choose $r_z\in\Bbb Q$. Now since the rationals are dense in the reals, the open ball $B(z, \frac {r_z}{10})$ contains a point with rational coordinates, call it $y$. Now it is easy to check that $$z\in B(y,\frac {r_z}{5})\subset B$$