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This question is inspired by https://math.stackexchange.com/a/1052384/66307 and quotes from it heavily.

Take a countably infinite paint box; this means that it has one color of paint for each positive integer; we can therefore call the colors $C_1, C_2, $ and so on. Take the set of real numbers, and imagine that each real number is painted with one of the colors of paint.

Now ask the question: Are there three distinct real numbers $a,b,c$, all painted the same color, such that $$a+b=c$$

Must such $a,b,c$, not all zero, exist regardless of how cleverly the numbers are actually colored?

  • What is your question? Are you asking if the problem you state is provable or do you want the answer? – 5xum Dec 08 '14 at 10:22
  • @5xum I would like to know if it is true and if so, how to prove it. Sorry if this wasn't clear. –  Dec 08 '14 at 10:23
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    The question is, in a nutshell, "Suppose that $f\colon\Bbb R\to\Bbb N$, are there [distinct] $a,b,c\in\Bbb R$ such that $a+b=c$ and $f(a)=f(b)=f(c)$?" (I've added distinct because taking $a=b=c=0$ works, otherwise). – Asaf Karagila Dec 08 '14 at 10:39

2 Answers2

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The paper I cited in the earlier question addresses precisely this matter:

There is an $\aleph_0$-coloring of the nonzero real numbers without a monochromatic solution to the Schur equation $x+y=z$ in distinct variables.

The coloring is very simple. First we color the positive reals so that there is no monochromatic solution in positive reals: simply color each $x$ with the color $\lfloor\log_2 x\rfloor$. The number $x$ is colored with color $i$ if and only if $x$ lies in the interval $[2^i, 2^{i+1})$. Then if $x$ and $y$ are both in $[2^i, 2^{i+1})$ their sum is in $[2^{i+1}, 2^{i+2})$ and so is colored with color $i+1$.

Now renumber the colors of the positive reals so that instead of being all integers, they are all the even integers except zero. Color the negative reals similarly using the odd integers: if $x$ is positive and has color $2i$, then $-x$ gets color $2i-1$. Finally, color $0$ with color $0$.

MJD
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It is at least consistent that the answer be no.

Assume $\mathsf{CH}$, and let $\{x_\xi:\xi<\omega_1\}$ be an enumeration of $\Bbb R$. For $\eta<\omega_1$ let $X_\eta$ be the closure of $\{x_\xi:\xi<\eta\}$ under addition and subtraction, and observe that $X_\eta$ is countable. For $\eta<\omega_1$ let $D_\eta=X_\eta\setminus\bigcup_{\xi<\eta}D_\xi$, and let $\mathscr{D}=\{D_\eta:\eta<\omega_1\}$; $\mathscr{D}$ is a partition of $\Bbb R$ into countable sets. For each $\eta<\omega_1$ let $f_\eta:D_\eta\to\omega$ be an injection, and let $f=\bigcup_{\eta<\omega_1}f_\eta$.

Suppose that $a,b,c\in\Bbb R$ are distinct, and that $a+b=c$. Let $\eta=\min\{\xi<\omega_1:a,b,c\in X_\xi\}$; then $|\{a,b,c\}\cap X_\xi|\le 1$ for each $\xi<\eta$. It follows that $|\{a,b,c\}\cap D_\eta|\ge 2$ and hence that $\{a,b,c\}$ is not monochromatic.

Brian M. Scott
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  • Thank you for this. So the question is, is it consistent that the answer is yes? –  Dec 09 '14 at 11:25