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.

  • $\begingroup$ 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. $\endgroup$ – lulu Jul 28 '16 at 23:58
  • $\begingroup$ 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 $\cup_{i\in I} B_i$. $\endgroup$ – Gregory Grant Jul 29 '16 at 0:01
  • 1
    $\begingroup$ @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? $\endgroup$ – lulu Jul 29 '16 at 0:04
  • 1
    $\begingroup$ $C$ 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. $\endgroup$ – Gregory Grant Jul 29 '16 at 0:11
  • $\begingroup$ Hint for the proof (closed ball case): Baire category theorem. $\endgroup$ – GEdgar Jul 29 '16 at 0:24

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.

  • $\begingroup$ Now, I'll try solve the Exercise :D $\endgroup$ – Oddone Jul 29 '16 at 0:50

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 $C'$).

  • $\begingroup$ This is very nice! $\endgroup$ – Noah Schweber Jul 29 '16 at 0:48
  • 1
    $\begingroup$ 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 $U\cap C'\not \subset \partial B_i.$ $\endgroup$ – DanielWainfleet Jul 30 '16 at 5:25
  • $\begingroup$ More details may be found in math.stackexchange.com/questions/2413138/… $\endgroup$ – GEdgar Sep 2 '17 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.