Show that every open subset can be represented as a union of at most countably many balls Consider $\mathbb{R^n}$ with the $\infty$-norm.
Show that every open subset can be represented as a union of at most countably many balls.
This might have been answered before but I do not understand the answers on there.
I have an idea that it needs to use the fact that the rationals are dense but I'm not sure how to make it rigorous.
I want to avoid using metric space notation but balls are fine.
 A: To generalize a little bit using the fact that the rationals are dense is the right idea:
Let $X$ be a metric space with a countably dense subset $A$. We will show that it has a countable basis.
Let $\mathbb{B}=\{b(a,1/n) \mid  a \in A, n \in \mathbb{Z_{+}}\}$.
Clearly, $\mathbb{B}$ is countable.
Let $x \in X$. Let $U$ be a neighborhood of $X$. Then there is an open ball $B(x,\delta) \subseteq U$. Then by the archimedean property, choose $N$ so that $1/n<\delta/2$. Since $A$ is dense, $x \in \overline{A}$. Then $B(x,1/n)$ intersects $A$ at some point $a$. Then $d(x,a)<1/n$ so $x \in B(a,1/n)$. 
It is then clear that $x \in B(a,1/n) \subseteq B(x,\delta) \subseteq U$, so $\mathbb{B}$ is a countable basis.

edit: a basis is a collection of open subsets of $X$ (or open intervals of $\mathbb{R}$ such that each $x \in X$ is contained in one of the basis elements. Another criteria, is that for each open ball about $x$, there is some basis element $B$ such that $x \in B \subseteq B(x,\epsilon)$. so pretty much, what is being shown is that for each $r \in \mathbb{R}$, you can find some $B(q, 1/n)$ where $q \in \mathbb{Q}$ and $n \in \mathbb{N}$ such that $x \in B(q, 1/n) \subseteq U$, where $U$ is an open interval containing $x$. Then $ r \in \mathbb{R} \implies r \in B$ for some $B \in \mathbb{B}$ and since there are only countably many $B$, any open subset of $\mathbb{R}$ is also a subset of some $B_{1} \cup B_{2}....$
Edit, part II:
Closure: let $A \subseteq \mathbb{R}$. $\overline{A}$ is the collection of points where any open ball about $r \in \mathbb{R}$ intersects $A$ at some point other than itself.
Dense: Let $A \subseteq \mathbb{R}$. if $r \in \mathbb{R}$, then $r \in \overline{A}$.
lemma: Let $r \in \mathbb{R}$. If there is some sequence of points $\{a_{n}\} \in A$ converging to $r$, then $r \in \overline{A}$.
So, in my sentence above, when I say $x \in \overline{A}$ implies that there is some $a \in A$ such that $a \in A \cap B(x,1/n)$, can you prove this with the lemma I provided?
So: for every $r \in \mathbb{R}$ you can find some sequence of $\{q\}$ in the rationals such that this sequence converges to $r$. 
Then $\mathbb{Q}$ is dense in $\mathbb{R}$.
A: Let $U$ be open in $\mathbb R^n.$ For every $x\in U,$ there exists $q_x\in \mathbb Q^n$ and $r_x\in \mathbb Q$ such that $x\in B(q_x,r_x)\subset U.$ Therefore
$$U = \cup_{x\in U} \{x\} \subset \cup_{x\in U} B(q_x,r_x) \subset U.$$
Therefore $U = \cup_{x\in U} B(q_x,r_x),$ and there are at most countably many distinct $B(q_x,r_x).$
A: I'm going to add this as a separate answer, but it is more or less the same answer, phrased differently. 
Suppose that $\mathbb{Q}$ is dense in $\mathbb{R}$. Then for each $r \in \mathbb{R}$ there is some sequence of $\{q_n\}$ in the rationals such that $\{q_{n}\}_{n=1}^{\infty}=r$.
Let $\mathbb{B}=\{B(q,1/n) \mid q \in \mathbb{Q}, n \in \mathbb{N}\}$.
Clearly $\mathbb{B}$ is a countable collection of open balls. It is also clear that $\mathbb{B} \subseteq \mathbb{R}$. We will show that $\mathbb{R} \subseteq \mathbb{B}$, showing that $\mathbb{R}=\bigcup B_{n}$ where $B_{n} \in \mathbb{B}$. It follows that any subset of the reals is also a union of open balls.
Suppose that $r \in \mathbb{R}$. Then $r \in B(r,\delta)$. Then by the archimedean property, there exists $n \in \mathbb{N}$ such that $1/n<\delta/2$.
Well, since there exists  $\{q_{k}\}_{k=1}^{\infty}=r$, we have that there is some $N \in \mathbb{N}$ such that $k>N \implies |q_{k}-r|<1/n$. Then there is some $q_k \in B(r,1/n)$. Then since $|r-q_n|<1/n$, $r \in B(q_n, 1/n)$. Then $r \in \bigcup B(q,1/n)=\mathbb{B}$, meaning that $\mathbb{R} \subseteq \mathbb{B}$. Thus $\mathbb{R}=\mathbb{B}$, a countable union of open balls.
