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I was trying to partition $\mathbb R$ into two sets $A, B$ such that for all $a\in A, b\in B$ we have $|a-b|\neq 1$. An obvious way to do it is to take $\mathbb Z$ and ${\mathbb R}\setminus {\mathbb Z}$. The other examples I found all consisted of one countable set and its complement.

Question. Is there $A\subseteq {\mathbb R}$ such that both $A$ and $B:= {\mathbb R}\setminus A$ are uncountable, and for all $a\in A, b\in B$ we have $|a-b|\neq 1$?

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    $\begingroup$ 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 $B := \{n+x\colon x \in [0,1)\setminus X, n \in \mathbb{Z}\}$. $\endgroup$
    – Sam Hopkins
    Commented Jan 3, 2023 at 17:20
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    $\begingroup$ 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.) $\endgroup$
    – YCor
    Commented Jan 3, 2023 at 17:30
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    $\begingroup$ (Sam Hopkins' description also provides all examples, slightly differently stated.) $\endgroup$
    – YCor
    Commented Jan 3, 2023 at 17:34
  • $\begingroup$ Thanks @SamHopkins and YCor - could you put your idea in an answer, Sam, so that we can close this thread? $\endgroup$
    – Dominic van der Zypen
    Commented Jan 3, 2023 at 19:50

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Converting my comment to an answer:

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 $A := \{n+x\colon n\in\mathbb{Z}, x\in X\}$ and $B := \{n+x\colon n \in \mathbb{Z}, x \in [0,1)\setminus X\}$.

It is easy to see that all solutions are of this form.

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