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world:boule [2025/10/03 01:56]
rdouc created
world:boule [2025/10/03 02:00] (current)
rdouc [Remarks:]
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-**Question** 
  
 +
 +====== Question ======
 +
 +
 +<WRAP center round tip 80%>
 Let $D$ be an infinite set without accumulation points in $\mathbb{R}^d$.  ​ Let $D$ be an infinite set without accumulation points in $\mathbb{R}^d$.  ​
 Show that, for every $n \in \mathbb{N}^*$,​ there exists a ball containing exactly $n$ points of $D$.  ​ Show that, for every $n \in \mathbb{N}^*$,​ there exists a ball containing exactly $n$ points of $D$.  ​
  
-**Proof**+</​WRAP>​
  
-First, note that $D$ must be countable. Indeed, for each point $x \in D$, one can associate a *rational ball* (that is, a ball with rational center coordinates and rational radius) containing $x$ as the only element of $D$. This defines an injection from $D$ into the countable set of rational balls, which proves that $D$ is countable.+===== Proof ===== 
 + 
 + 
 +First, note that $D$ must be countable. Indeed, for each point $x \in D$, one can associate a **rational ball** (that is, a ball with rational center coordinates and rational radius) containing $x$ as the only element of $D$. This defines an injection from $D$ into the countable set of rational balls, which proves that $D$ is countable.
  
 Now consider, for $(a,b) \in D \times D$, the affine hyperplane Now consider, for $(a,b) \in D \times D$, the affine hyperplane
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-**Remarks:**+ 
 +<WRAP center round important 80%> 
 +===== Remarks: ​=====
  
   * One can avoid Baire’s theorem by using a measure-theoretic argument. Let $\lambda_d$ be the Lebesgue measure on $\mathbb{R}^d$. Since $\lambda_d(H_{a,​b})=0$,​ we get $\lambda_d(\bigcup_{a,​b \in D} H_{a,​b})=0$. Hence $\bigcup_{a,​b \in D} H_{a,b} \neq \mathbb{R}^d$.   * One can avoid Baire’s theorem by using a measure-theoretic argument. Let $\lambda_d$ be the Lebesgue measure on $\mathbb{R}^d$. Since $\lambda_d(H_{a,​b})=0$,​ we get $\lambda_d(\bigcup_{a,​b \in D} H_{a,​b})=0$. Hence $\bigcup_{a,​b \in D} H_{a,b} \neq \mathbb{R}^d$.
   * For $D = \mathbb{Z}^d$,​ one can construct $c$ explicitly, for example $c=(\pi,​\ldots,​\pi^d)$. Then, for $a,b \in \mathbb{Z}^d$,​ the equality $\|c-a\|^2=\|c-b\|^2$ becomes a polynomial equation with integer coefficients having $\pi$ as a root. This is impossible unless $a=b$, since $\pi$ is not algebraic. However, this explicit construction of $c$ does not generalize (or at least not simply) when $D$ is not $\mathbb{Z}^d$.   * For $D = \mathbb{Z}^d$,​ one can construct $c$ explicitly, for example $c=(\pi,​\ldots,​\pi^d)$. Then, for $a,b \in \mathbb{Z}^d$,​ the equality $\|c-a\|^2=\|c-b\|^2$ becomes a polynomial equation with integer coefficients having $\pi$ as a root. This is impossible unless $a=b$, since $\pi$ is not algebraic. However, this explicit construction of $c$ does not generalize (or at least not simply) when $D$ is not $\mathbb{Z}^d$.
   * Obviously, when $D = \mathbb{Z}^d$,​ one does not need to prove that $D$ is countable, nor that $A \cap [0,r]$ is finite, which significantly shortens the proof.   * Obviously, when $D = \mathbb{Z}^d$,​ one does not need to prove that $D$ is countable, nor that $A \cap [0,r]$ is finite, which significantly shortens the proof.
 +
 +</​WRAP>​
  
world/boule.1759449407.txt.gz · Last modified: 2025/10/03 01:56 by rdouc