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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.

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    Your idea for (1) is good, you can define a surjective map $\mathbb Q^n \times \mathbb Q_+$ to your collection. Thus the collection is countably infinite. For (2). you need to show: for all open set $V$ and $x\in V$, there is $B \in \mathcal{B}$ so that $x\in B\subset V$. The idea is to choose $q in V$ closed to $x$ so that a small ball centered at $q$ with rational radius contains $x$ and is still inside $V$. – 2017-02-08
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    Since Q is dense in R. Isn't it a given that for any x \in R there will be a y \in Q closer than any epsilon? So doesnt a proof follow from that? Or is there more to it? – 2017-02-08
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    Yes, more or less. You just need to be a bit careful in choosing (1) $\epsilon$ is rational and (2) the $\epsilon$-ball is in $V$ and still is contained $V$. – 2017-02-08

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(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.

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    "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" Can you elaborate on this part? I think I know what you mean, but I'm not clear. Say we have a distance $d$ from v to the border of $B$, but if that distance is irrational, how do we derive a rational radius from it? – 2017-02-08
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    You have a minimum distance $d_1$ from $v$ to the border of $B$. Choose a $w$ with the distance $d_2$ from $v$ to $w$ strictly less than $\frac{d_1}{2}$. Then if you choose a radius $r\in(d_2, \frac{d_1}{2})$, the open ball must be inside $B$ and must contain $v$. The rationals are dense in $\mathbb{R}$, so there exists a rational in that interval. – 2017-02-08
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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.

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    Not really, no. – 2017-02-08
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    Oh, is it when it satisfies (2) above? As if in any open set in A can be written as a countable union of members of B, then B is a basis for A? This question introduced me to the concept, but I am not too familiar with it. – 2017-02-08
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    @jackson5 I added some definitions, and gave an alternative way to find a countable basis. Did this clear anything up for you? – 2017-02-08