world:boule
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world:boule [2025/10/03 01:58] rdouc |
world:boule [2025/10/03 02:00] (current) rdouc [Remarks:] |
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| - | 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. | + | 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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| + | <WRAP center round important 80%> | ||
| ===== Remarks: ===== | ===== Remarks: ===== | ||
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| * 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. | ||
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| + | </WRAP> | ||
world/boule.1759449495.txt.gz ยท Last modified: 2025/10/03 01:58 by rdouc
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