# Every open subset of $\mathbb{R}^{p}$ is the union of a countable collection of closed sets.

I want to prove the following statement which is in The Elements of Real Analysis by Robert G. Bartle, on page 67 exercise H:

Every open subset of $\mathbb{R}^{p}$ is the union of a countable collection of closed sets. (Hint: Argue as in the preceding exercise, but this time use closed balls)

I have referred to the to the solution given in this question, but the approach is different from what the book is asking for. I first show the preceding exercise and its proof, and then I show my attempt at the statement that I want to prove.

Preceding Exercise: A subset of $\mathbb{R}^{p}$ is open iff it is the union of a countable collection of open balls.

proof. $(\leftarrow )$ An open ball is an open set, and by property 9.3(c) the union of any collection of open sets in $\mathbb{R}^{p}$ is indeed open in $\mathbb{R}^{p}$. That is that the union of a countable collection of open balls is open in $\mathbb{R}^{p}$, hence it is an open subset of $\mathbb{R}^{p}$.

$(\rightarrow)$ Let $G\neq\emptyset$ , $G\subset\mathbb{R}^{p}$ be open, let $\{r_{n}\::\: n\in\mathbb{N}\}$ be an enumeration of all the rational points in $G$ (that is that each of the $p$ components of a point of $G$ , consisting of only rational numbers). Now we let $\mathcal{B}(r_{n},\frac{1}{m_{n}}$) where $m_{n}$ is the smallest natural number such that $\mathcal{B}(r_{n},\frac{1}{m_{n}})\subset G$ . Clearly,$$\bigcup_{n\in\mathbb{N}}\mathcal{B}(r_{n},\frac{1}{m_{n}})\subset G$$.

Now if we let $x\in G$ , since $G$ is open we know that there exists a ball centered at $x$ , choose $m$ such that $\mathcal{B}(x,\frac{2}{m})\subset G$ . Now it follows from Theorem 6.10 that there exists a rational number $y\in\mathcal{B} (x,\frac{1}{m})$ ; thus $y\in G$ and therefore $y=r_{n}\in\{r_{n}\::\: n\in\mathbb{N}\}$ for some $n$ . If $x\notin\mathcal{B}(r_{n},\frac{1}{m_{n}})$ then we must have that $\frac{1}{m_{n}}<\frac{1}{m}$ . [(why?) $y\in\mathcal{B}(x,\frac{1}{m})$ implies that the distance $d(x,y)<\frac{1}{m}$ , while $x\notin\mathcal{B}(r_{n},\frac{1}{m_{n}})=\mathcal{B}(y,\frac{1}{m_{n}})$ implies that the distance $d(x,y)\geqslant\frac{1}{m_{n}}$ ; hence $\frac{1}{m_{n}}<\frac{1}{m}$ ]. Now, clearly $\mathcal{B}(r_{n},\frac{1}{m})\subset\mathcal{B}(x,\frac{2}{m})\subset G$ which contradict our selections on $m_{n}$ , thus $x\in\mathcal{B}(r_{n},\frac{1}{m_{n}})$ . [(again,why?) Let $\alpha\in\mathcal{B}(r_{n},\frac{1}{m})$ , then $d(\alpha,y)<1/m$ it follows from the triangle inequality that $d(x,\alpha)\leqslant d(x,y)+d(y,\alpha)<\frac{2}{m}$ ; hence $\alpha\in\mathcal{B}(x,\frac{2}{m})$ and therefore $\mathcal{B}(r_{n},\frac{1}{m})\subset\mathcal{B}(x,\frac{2}{m})$ ]. Now, since $x$ was arbitrarly in $G$ , it follows that for each $x\in G$ there exists $n$ such that $x\in\mathcal{B}(r_{n},\frac{1}{m_{n}})$ ; hence $G\subset\bigcup_{n\in\mathbb{N}}\mathcal{B}(r_{n},\frac{1}{m_{n}})$ which shows that $G=\bigcup_{n\in\mathbb{N}}\mathcal{B}(r_{n},\frac{1}{m_{n}})$. $\square$

Proposition to be proved:

Every open subset of $\mathbb{R}^{p}$ is the union of a countable collection of closed sets.

proof. Let $G\ne\emptyset$ , $G\subset\mathbb{R}^{p}$ be open, let $\{r_{n}\::\: n\in\mathbb{N}\}$ be an enumeration of all the rational points in $G$ . Since, $G$ is open for every $r_{n}$ there exists an open ball $\mathcal{B}(r_{n},\varepsilon_{n})$ contained in $G$ . Moreover, for each $r_{n}$ there exists a closed ball $\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})\subset\mathcal{B}(r_{n},\varepsilon_{n})\subset\begin{subarray}{c} G\end{subarray}$ , where $m_{n}$ is the smallest natural number such that $\varepsilon_{n}-\frac{1}{m_{n}}>0$ . (i.e. $\frac{1}{m_{n}}<\varepsilon_{n}$ ). Clearly,

$$\bigcup_{n\in\mathbb{N}}\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})\subset G.$$

Now we pick an arbitrary $x\in G$ , since $G$ is open we can find a $\mathcal{B}(x,\varepsilon_{m})\subset G$; choose $m\in\mathbb{N}$ so that $m$ is the smallest number such that $\varepsilon_{m}-\frac{1}{m}>0$ and $\overline{\mathcal{B}}(x,\varepsilon_{m}-\frac{1}{m})\begin{subarray}{c} \subset\end{subarray}\mathcal{B}(x,\varepsilon_{m})$ . Now, since $\varepsilon_{m}-\frac{1}{m}>0$ it follows that there must exist a rational number $y\in\overline{\mathcal{B}}(x,\varepsilon_{m}-\frac{1}{m})$ . [hence $y=r_{n}$ for some $n$ and $\mathcal{\overline{B}}(y,\varepsilon_{n}-\frac{1}{m_{n}})=\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})\subset\mathcal{B}(x,\varepsilon_{m})$ ]. Suppose that $x\notin\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})$ , then $d(x,y)>\varepsilon_{n}-\frac{1}{m_{n}}$ , but we know $d(x,y)\leqslant\varepsilon_{m}-\frac{1}{m}$ so it follows that $0<\varepsilon_{n}-\frac{1}{m_{n}}<\varepsilon_{m}-\frac{1}{m}$ . Now, let $\alpha\in\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})$ it follows that $d(\alpha,y)\leqslant\varepsilon_{n}-\frac{1}{m_{n}}$ and that $d(x,\alpha)\leqslant d(\alpha,y)+d(y,x)\leqslant\left(\varepsilon_{n}-\frac{1}{m_{n}}\right)+\left(\varepsilon_{m}-\frac{1}{m}\right)$.

As you might see by now I am spinning my wheels and getting no where! If I could show that $x\in\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})$ then I would be done, or likewise if I could show $\overline{\mathcal{B}}(r_{n},\varepsilon_{n}-\frac{1}{m_{n}})\subset\overline{\mathcal{B}}(x,\varepsilon_{m}-\frac{1}{m})$ but I can't do either of these, because I don't have any way to relate $\varepsilon_{n}$ to $\varepsilon_{m}$. Perhaps my approach is entirely wrong. I would greatly appreciate some guidance, thanks!

• You got it. Start it over on a fresh empty paper. – Berci Dec 1 '13 at 20:56
• @Berci I don't see how I have it? I haven't used the fact that I chose $\frac{1}{m_n}$ to be the smallest natural number such that $\frac{1}{m_n}<\varepsilon_n$ yet, do I even need to assume this? – JimmyJackson Dec 1 '13 at 21:02

You are working entirely too hard. You already know that every open subset of $\Bbb R^n$ is the union of a countable collection of open balls, and I’m sure that you know that the union of a countable family of countable collections is still countable. Thus, all you have to do is show that every open ball in $\Bbb R^n$ is the union of a countable collection of closed sets. In fact you can show that every open ball in $\Bbb R^n$ is the union of a countable collection of closed balls, as the hint suggests. In all other respects, however, the hint is counterproductive, because there is no need to repeat the complexities of the argument for the earlier exercise.
Let $B(p,r)$ be the open ball in $\Bbb R^n$ of radius $r$ centred at $p$. Can you find a very simple countable family of closed balls whose union is $B(p,r)$? Think about it for a while; if you don’t see it, there’s a big hint in the spoiler-protected block below.