Timeline for Partitioning $\mathbb R$ into sets such that no mutual points have distance $1$ [closed]
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2023 at 6:20 | history | closed |
Arno YCor Loïc Teyssier Piotr Hajlasz Andreas Blass |
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| Jan 3, 2023 at 20:01 | vote | accept | Dominic van der Zypen | ||
| Jan 3, 2023 at 19:55 | answer | added | Sam Hopkins | timeline score: 6 | |
| Jan 3, 2023 at 19:50 | comment | added | Dominic van der Zypen | Thanks @SamHopkins and YCor - could you put your idea in an answer, Sam, so that we can close this thread? | |
| Jan 3, 2023 at 19:08 | review | Close votes | |||
| Jan 4, 2023 at 6:27 | |||||
| Jan 3, 2023 at 17:34 | comment | added | YCor | (Sam Hopkins' description also provides all examples, slightly differently stated.) | |
| Jan 3, 2023 at 17:31 | history | edited | YCor |
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| Jan 3, 2023 at 17:30 | comment | added | YCor | The group $\mathbf{R}$ is the disjoint of the cosets of the subgroup $\mathbf{Z}$. So we need a partition which on every coset, satisfies the given condition. On a coset, the condition means that either $A$ or $B$ is empty. So the solutions are exactly the unions of cosets of $\mathbf{Z}$, i.e., the subsets of $\mathbf{R}$ that are invariant under integral translations, i.e., the inverse images $p^{-1}(Y)$ of subsets $Y$ of the circle $\mathbf{R}/\mathbf{Z}$ under the canonical projection $p:\mathbf{R}\to\mathbf{R}/\mathbf{Z}$. (In particular, there are $2^c$ such partitions.) | |
| Jan 3, 2023 at 17:20 | comment | added | Sam Hopkins | Choose any $X \subseteq [0,1)$ which is uncountable and for which $[0,1)\setminus X$ is also uncountable (for example $X=[0,\frac{1}{2}]$); then set $A := \{n+x\colon x \in X, n \in \mathbb{Z}\}$ and $B := \{n+x\colon x \in [0,1)\setminus X, n \in \mathbb{Z}\}$. | |
| Jan 3, 2023 at 17:15 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |