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Does not exist cover of $\mathbb{R}^n$ by disjoint closed balls with positive radius.

My attempt: Suppose that exists, we can write:

$\mathbb{R}^n=\displaystyle\bigcup_{i=1}^{\infty} B_{i}$. Let $C$ denote the set of center points of these balls. All points of $C$ are isolated so $C$ is countable. Associating each point of $C$ to the ball of wich this point is the center,it is easy see that this association is a bijection, concluding that set of balls is countable.

The set of limit points of $C$, $C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i}$. I thought it would be easy to find a contradiction there, I was wrong. Until now, I don't use hypothesis that $B_{i}$ is closed, and I think that is it that lack in my demostration.

Is it true if the balls are open?

Thank you for any help.

Oddone
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  • Open balls won't work: Take a set $U$ in your cover, then the union of all the other sets in the cover (excluding $U$) is open so you've disconnected $\mathbb R^n$, a contradiction. – lulu Jul 28 '16 at 23:58
  • You are assuming the cover is necessarily countable, but a cover need not be countable. Instead you need an arbitrary index set $I$ and write $\cup_{i\in I} B_i$. – Gregory Grant Jul 29 '16 at 00:01
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    @GregoryGrant Can it be uncountable? If you had such a covering then, for each $B$ in your collection we could find an element $x_B\in \mathbb Q^n\cap B$ thereby giving an injection from your collection to $\mathbb Q^n$...or have I got that wrong? – lulu Jul 29 '16 at 00:04
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    $C$ is countable only because you assumed the cover is countable. If the question is about an arbitrary cover you need to first prove you can reduce to the case of a countable cover. A space where every cover has a countable subcover is called a Lindelöf space and indeed $\Bbb R^n$ is Lindelöf, but you need to at least say that to reduce to the countable cover case, you can't just start with a countable cover with no justification. – Gregory Grant Jul 29 '16 at 00:11
  • Hint for the proof (closed ball case): Baire category theorem. – GEdgar Jul 29 '16 at 00:24

2 Answers2

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For the Baire category proof.
Use the OP proof up to $$ C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i} . $$ Then note $C^{'}$ is a complete metric space, written as a countable union of sets $\partial B_i$, which are closed (in $C'$) sets with empty interior (in $C'$).

GEdgar
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  • This is very nice! – Noah Schweber Jul 29 '16 at 00:48
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    Beautiful!..... It may not be obvious that $C'$ is a complete metric space : But $C'$ is actually closed because any limit point of $C'$ cannot belong to $int(B_i)$ for any $i$ so it must belong to $R^n$ \ $\cup_i int(B_i)=C'.$ ....It may not be obvious that $\partial B_i$ has empty interior in $C'$ : But for $ p\in \partial B_i,$ any open $U$ of $R^n$ with $p\in U$ satisfies $U\cap B_j \ne \emptyset$ for some $j\ne i,$ and the closest member of $B_j \cap U$ to $p$ must belong to $\partial B_j.$ So $U\cap C'\not \subset \partial B_i.$ – DanielWainfleet Jul 30 '16 at 05:25
  • More details may be found in https://math.stackexchange.com/questions/2413138/question-about-nowhere-dense-condition-in-baire-category-theorem-application# – GEdgar Sep 02 '17 at 11:58
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To show no such cover via open balls exists is easy: any such cover $\mathcal{O}$ must involve at least 2 open balls (since the radii are finite). So pick $O\in\mathcal{O}$, and let

  • $A=O$,

  • $B=\bigcup (\mathcal{O}\setminus\{O\})$.

Then $A$ is open by assumption, while $B$ is open as a union of open sets; and $A\sqcup B=\mathbb{R}^n$. But this contradicts the connectivity of $\mathbb{R}^n$.

Here's one way to do it. Suppose towards contradiction that $\mathcal{U}$ is a set of countably many disjoint closed balls with finite positive radius, which covers $\mathbb{R}^n$ (you've justified in your question why we may assume the cover is countable). Say that a closed ball $B$ is shattered if it is not contained in a single element of $\mathcal{U}$.

  • Exercise: there is a shattered closed ball.

Now the crucial point is:

If $B$ is a shattered closed ball, and $U\in\mathcal{U}$, then there is a closed ball $C$ of aribtrarily small positive radius contained in $B\setminus U$ which is also shattered.

Why? Well, clearly there are $x, y\in B\setminus U$ which lie in different elements $V_x, V_y$ of $\mathcal{U}$ (why?). This means that $\partial V_x\cap (B\setminus U)$ is nonempty (why?); let $z$ be an element of this boundary. Then an appropriately small-radius closed ball around $z$ is shattered.

OK, so what? Well, now we can diagonalize against $\mathcal{U}$! Let $\mathcal{U}=\{U_i: i\in\mathbb{N}\}$, and define a sequence of shattered closed balls of finite positive radius $C_i$ such that

  • $C_{i+1}\subseteq C_{i}\setminus U_i$, and

  • The diameter of $C_i$ is at most $2^{-i}$.

But then $\bigcap C_i=\{\alpha\}$ for some $\alpha\in\mathbb{R}^n$, which can't be an element of any element of $\mathcal{U}$ (why?); contradiction.

Noah Schweber
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