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world:boule [2025/10/03 08:46] rdouc [Proof] |
world:boule [2025/10/03 09:06] (current) rdouc [Proof] |
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| - | To order these distances strictly, it is now enough to check that, for every $r>0$, the set $A \cap [0,r]$ is finite. But $A \cap [0,r]$ corresponds to the number of points in $D \cap B(c,r)$, which is a compact set containing no accumulation points. Hence it is finite. | + | To order these distances strictly, it is now enough to check that, for every $r>0$, the set $A \cap [0,r]$ is finite. But the cardinal of $A \cap [0,r]$ corresponds to the number of points in $D \cap \overline{B(c,r)}$, which is a compact set containing no accumulation points. Hence it is finite. |
| - | We can thus write $A = \{\rho_n \; ; \; n \geq 1\}$ with $\rho_n$ strictly increasing. | + | We can thus write $A = \{\rho_n \; ; \; n \geq 1\}$ with strictly increasing $(\rho_n)$. |
| For every $n \geq 1$, by choosing a radius $\rho$ such that $\rho_n < \rho < \rho_{n+1}$, the ball $B(c,\rho)$ contains exactly $n$ points of $D$. | For every $n \geq 1$, by choosing a radius $\rho$ such that $\rho_n < \rho < \rho_{n+1}$, the ball $B(c,\rho)$ contains exactly $n$ points of $D$. | ||
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| ===== Remarks: ===== | ===== 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 $\lambda_d(H_{a,b})=0$ and $D$ is countable, 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. | ||
world/boule.1759473966.txt.gz · Last modified: 2025/10/03 08:46 by rdouc
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