Open sets and rational sets (Topology)

Question

Let be $U\subset R^n$ an not empty open set and $\mathbb{Q}^n$ (i.e. $\underbrace{\mathbb{Q}\times\dotso\times\mathbb{Q}}_{\text{n times}}$)

Proof that $U$ can be written like the union of balls with center in $\mathbb{Q}^n$ and rational radius.

I have already proven that $U \cap \mathbb{Q}^n \neq \emptyset $ but I am not sure how to get that rational radius.

Answer 1

It might be easier to visualize the case $n=1$.

Let $x \in U$. Choose $\varepsilon$ small enough so that ball, $B_\varepsilon(x):=(x-\varepsilon, x+\varepsilon)$ of radius $\varepsilon $ cetnered at $x$ is contained in $U$. (You should be clear what it means to be open.) Prove:

1. For each $y \in B_\varepsilon(x)$ show that exists $r \in Q$, $y\in B(r(y))$ where $B(r(y)) \subseteq B_\varepsilon(x)$ is centered at $r$.

For example $\pi \in (1,4)$, then there exists a ball centered at $3$ containing it.

2. Show that $B(r)$ maybe be taken of rational radius.

3. Show $B_\varepsilon(x) = \bigcup_{y \in B_\varepsilon(x)} B(r(y))$ and $U=\bigcup_x B_{\varepsilon(x)}(x)$.

4. Deduce by combining 3 and 4.

Generalize above for other dimensions (simply replace the intervals). You have to be clear what it means to be open in $\mathbb{R}^n$.

For 3. Union implies each $y \in LHS$ is in $RHS$. and by construction too, each $B(r(y)) \subseteq LHS$.