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Is that possible to partition $(\Bbb R,+)$ into 4 additively closed subsets?

It is easy to partition it to 1 and 2 and 3 additively closed subsets. Fore example for 2 subsets we have: $(\Bbb R^{\geq 0},+)\cup(\Bbb R^{< 0},+)$. But Is that possible for 4 additively closed subsets? What about $k$ additively closed subsets?

C.F.G
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2 Answers2

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You can partition $(\Bbb R^n,+)$ into the three subsemigroups $H^n_+=\{x\in\Bbb R^n\,:\, x_n>0\}$, $H^n_-=\{x\in\Bbb R^n\,:\,x_n<0\}$ and $\Bbb R^{n-1}\times\{0\}$. Therefore, if $\Bbb R^{n-1}$ can be partitioned into $k$ subsemigroups, then $\Bbb R^n$ can be partitioned into $k+2$ subsemigroups. Now, let $m$ be the smallest positive natural number such that $(\Bbb R,+)$ cannot be partitioned in $m$ subsemigroups. As you said, $m\ge3$, and by definition $(\Bbb R,+)$ can be partitioned in $m-2$ subsemigroups. But then $(\Bbb R^2,+)$ can be partitioned in $m$ subsemigroups. But thanks to AC (axiom of choice), $(\Bbb R,+)$ and $(\Bbb R^2,+)$ are isomorphic as groups, and therefore $(\Bbb R,+)$ can be partitioned into $m$ subsemgroups as well, against the hypothesis on $m$. Therefore $(\Bbb R,+)$ can be partitioned into any finite number of subsemigroups.

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    That's very cute! (Of course it leaves open the question of what happens if we drop choice ... :P) – Noah Schweber Aug 24 '20 at 16:10
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    That's really cool. Is it possible to explicitly write down a better partition, or do we really need AC? – Mathmo123 Aug 24 '20 at 16:11
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    @Mathmo123 One way to make that question more precise would be to ask if it's possible to come up with a partition into Borel sets (and if so, how high in the Borel hierarchy do they have to be?). I have no clue how to answer that question, but I'd be very interested if someone else has an idea. – Chris Eagle Aug 24 '20 at 16:32
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    Sorry, but what is the AC? (In electronic we have AC and DC current:)! also we have AC and BC :)!) – C.F.G Aug 24 '20 at 16:32
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    @C.F.G It's referring to the Axiom of Choice: https://en.wikipedia.org/wiki/Axiom_of_choice. – Chris Eagle Aug 24 '20 at 16:37
  • @ChrisEagle: thanks. Where can I read the proof of $(\Bbb R,+)\simeq(\Bbb R^2,+)$? Is it easy? – C.F.G Aug 24 '20 at 16:39
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    Its proof can be found here. – C.F.G Aug 24 '20 at 16:43
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    I was interested enough in the question about Borel partitions that I posted it as a separate question, here: https://math.stackexchange.com/questions/3801882/complexity-of-partitions-of-mathbbr-into-subsemigroups – Chris Eagle Aug 24 '20 at 16:56
  • @Mathmo123 and everyone interested in the necessity of AC: I don't know, I think you should ask an actual mathematician. –  Aug 24 '20 at 17:22
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    This argument requires AC, since it is consistent that $\Bbb R^2$ is not isomorphic to $\Bbb R$ as additive groups. Whether or not there is a partition which does not rely on the axiom of choice, I don't know. The first, and obvious, thing to check is whether or not any of the automatic continuity theorems (e.g. Pettis' theorem) will work for semigroups. If so, it seems reasonable that any such partition will have to be to connected semigroups, which means that it must be negative and positive. – Asaf Karagila Aug 24 '20 at 17:41
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    I think this statement fails without AC, as I think the sets involved cannot be Lesbegue measureable. It seems to follow from this MSE question: https://math.stackexchange.com/questions/3048432/partition-of-positive-reals-with-each-part-closed-under-addition-without-choice. – David A. Craven Aug 24 '20 at 18:59
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For completeness I’ll add an answer showing that in fact $\Bbb R$ can be partitioned into $\kappa$ sets closed under addition for any non-zero cardinal $\kappa\le 2^\omega=\mathfrak{c}$. (Of course this uses the axiom of choice.) Start with a Hamel basis $B=\{b_\xi:\xi<2^\omega\}$ for $\Bbb R$ over $\Bbb Q$. For each $x\in\Bbb R\setminus\{0\}$ there is a unique finite $B_x\subseteq B$ such that $x$ is a linear combination with non-zero rational coefficients of the members of $B_x$; let $B_x^+$ be the set of members of $B_x$ whose coefficients in that linear combination are positive. For each $\eta<2^\omega$ let

$$A_\eta=\big\{x\in\Bbb R:\min\{\xi<2^\omega:b_\xi\in B_x\}=\eta\text{ and }b_\eta\in B_x^+\big\}\,;$$

$b_\eta\in A_\eta$, so $A_\eta\ne\varnothing$, and $A_\eta$ is clearly closed under addition.

Now let $\kappa\le 2^\omega$ be a cardinal, and let $$D=\Bbb R\setminus\bigcup_{\xi<\kappa}A_\xi\,.$$

Clearly $\{D\}\cup\{A_\xi:\xi<\kappa\}$ is a partition of $\Bbb R$ into $\kappa$ parts if $\kappa\ge\omega$, and into $\kappa+1$ parts if $\kappa<\omega$, and it only remains to show that $D$ is closed under addition. But $x\in D$ iff either

  • $\min\{\xi<2^\omega:b_\xi\in B_x\}\ge\kappa$, or
  • $\eta=\min\{\xi<2^\omega:b_\xi\in B_x\}<\kappa$ and $b_\eta\notin B_x^+$, or
  • $x=0$,

and it’s easy to check that the set of real numbers satisfying one of these conditions is closed under addition.

Brian M. Scott
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