Prove that the collection of all rational $\epsilon$-neighbourhoods of $\mathbb{R}^n$ centered on $x \in \mathbb{Q}^n$ is countable

Question

Let $\mathcal{B}$ denote the collection of all $\epsilon$-neighborhoods in $\mathbb{R}^m$ whose radius $\epsilon$ is rational and whose center has all coordinates rational.

I'm tryin to prove that (1) $\mathcal{B}$ is countable and that (2) every open subset of $\mathbb{R}^m$ can be expressed as the countable union of members of $\mathcal{B}$ .

(1) I get that all epsilon balls in $\mathcal{B}$ are centered on members of $\mathbb{Q}^n$. Is it possible to proceed by saying that $\mathcal{B}$ has the same cardinality as $\mathbb{Q}^{n+1}$?? Because the additional coordinate could denote $\epsilon$.

(2) I intuitively understand why the second part is true but I am having difficulty proving it.

Answer 1

(1) To your first point, yes you can absolutely say that $\mathcal{B}$ has the same cardinality as $\mathbb{Q}^{n+1}$. One way to do so is using the very natural bijection you had in mind to make such a suggestion. Map the first $n$ coordinates of $v\in\mathbb{Q}^{n+1}$ to the center of an open ball in $\mathcal{B}$, and map the last coordinate to the radius. Since $\mathbb{Q}^{n+1}$ is countable, so is $\mathcal{B}$.

(2) A standard trick is to let an open ball $B\in\mathbb{R}^n$ be given. Then, if $v$ is a point in $B$, there exists some $b_v\in\mathcal{B}$ so that $b_v\subseteq B$ and $v\in b_v$. That open ball $B$ is then just the union $\bigcup_{v\in B}b_v$. Since every open set in $\mathbb{R}^n$ is some arbitrary union of open balls (or empty), and since every open ball is a union of elements in $\mathcal{B}$, it follows that every open set is a union of elements of $\mathcal{B}$. That union is countable since $\mathcal{B}$ is countable.

Demonstrating the existence of $b_v$ can be done by considering the minimum distance from $v$ to the boundary of $B$. There is a rational center less than half that distance from $v$, and you can choose a rational radius large enough to encompass $v$ but small enough to stay within $B$. You can use thirds or other smaller fractions instead of halves if you don't want to think about technical details.

Answer 2

For $n=1$, you can take the basis

$$\{B(q_k,1/n) : q_k \in \mathbb Q, n \in \mathbb N\}$$

Do you know how to tell when a collection is a topological basis? It will suffice to check that every real number can be written this way, and that if this real number is in two distinct open sets, there is a third that also contains it. Then this is a basis. To see that it generates the standard topology, check that for all $x \in U$, there is an intermediate set $B$ among the basis elements so that $x \in B \subset U$.

I think an alternative could be open boxes with rational end points, which would be a countable collection, while still generating all the open intervals, and hence open sets.