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I am having problems proving the following claim: Given a bounded set $A \subset R^n$, I want to prove the existence of $a_1, \dots, a_N \in R^n$ and numbers $r_1, \dots, r_N \in [0, +\infty)$ such that

$$A \subset \cup_{k=1}^N B(a_k,r_k)$$

Where the ball is defined ad

$$B(a,r) = \{ x \in R^n : |x - a| \leq r \}, a \in R^n, r \geq 0$$

and the set $\{\sum_{k=1}^N r^2_k : \text{A can be covered with a collection} \ B(a_k,r_k)_{k=1}^N\}$ has a smallest element (maybe the balls have to be closed, to be precised ...)

I think I should use Bolzano-Weierstrass theorem, but I am struggling with a formal proof.

If A is bounded that means I can construct a finite coverage by the union of finite balls centered at $a_k$. Do I need to consider the case of A closed and A open?

user186918
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  • Do you have a definition of compactness to the effect of "every open cover has a finite subcover"? If so, you should use it here. – Ben Grossmann Oct 24 '14 at 03:57
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    Maybe you should include the fact that $N$ needs to be fixed. Otherwise the statement is false. Consider A to be a line segment in $\mathbb{R}^2$. – Dunham Oct 27 '14 at 16:01

2 Answers2

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WLOG, suppose $A\subseteq [0,1]^n$, and define the collection of coverings \begin{equation*} \mathcal{C} = \left\{ (a_1,\dots,a_N)\times(r_1,\dots,r_N) \in [0,1]^n \times [0,2^n]^n: A\subseteq \bigcup_{k=1}^N B(a_k,r_k) \right\}. \end{equation*}

Part I: $\mathcal{C}$ is closed or equivalently $\mathcal{C}^c$ is open.

Take any point $E = (a_1,\dots,a_N)\times(r_1,\dots,r_N) \in \mathcal{C}^c$. Then there is a point $a_0 \in A$ such that $a_0 \notin B(a_k,r_k)$ for any $k$. For each $k$, $B(a_k,r_k)$ is closed, so we can perturb the ball by sufficiently small factors $0<\epsilon_{k}$, $\delta_{k}\in [-1,1]^n$ so that $a_0 \notin B(a_k+\delta_k,r_k+\epsilon_k)$. This shows $\mathcal{C}^c$ is open.

Part II:

For $C= (a_1,\dots,a_N)\times(r_1,\dots,r_N) \in \mathcal{C}$, define $\rho(C) = \sum_{k=1}^N r_k^2$, and let \begin{equation*} M := \inf_{C\in \mathcal{C}} \rho(C). \end{equation*} We now define a sequence $C_m\in \mathcal{C}$ such that $\rho(C_m)\leq M+1/m$ for $m\in\mathbb{N}$. Since $\mathcal{C}$ is a compact set, there is a subsequence $C_{m_l}$ of $C_m$ that converges to a point $C_0=(a_1^0,\dots,a_N^0)\times(r_1^0,\dots,r_N^0)$. By construction, $M=\rho(C_0)$.

Dunham
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There is an easy way to prove that if you consider the closure of A (who will be closed and bounded, hence compact) and the fact that A is included in its closure. You can then apply borel-lebesgue theorem to the closure of A and get this result.

Of course this proof depends on which definition of compacity you relie on, but you can prove equivalence between different definition if needed ( A compact iff from every sequence of element in A you can extract a converging one, or A compact iff there is a finite number of balls with a radius r such that their union includes A which is what you're looking for ).

If you want I can detail the proof involving the equivalency between these two definitions (in your case, that is A part of $R^n$), but you can look at this page of wikipedia at the "open cover definition" :

http://en.wikipedia.org/wiki/Compact_space#Open_cover_definition

mvggz
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  • the difficulty is not the existence of a covering, it is the question of a smallest covering. – Dunham Oct 27 '14 at 15:55
  • Yes I saw that, but my point was to use the closure of A and thus ease the proof. But yours is nice :) , and it's not certain that using what I said will do the trick – mvggz Oct 27 '14 at 16:02