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Is there a colouring of $\mathbb{R}^2$ with $2$ colours, such that the just single-coloured paths are constant ones? (Note: I've got this problem from another one.)

  • Do you mean that any continuous curve that only lies within a single colour is a straight line? A horizontal straight line? – Arthur Feb 26 '18 at 12:51
  • @Arthur Maybe using the word curve, makes someones to think that the straight lines are permitted. But I didn't mean this. I edited the post and used the word path, instead of it. –  Feb 26 '18 at 13:06
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    Oh, ok. So an actual curve $\gamma:[0,1]\to \Bbb R^2$, not the graph of a function $\Bbb R\to \Bbb R$ or anything. – Arthur Feb 26 '18 at 13:08
  • It can be done with 4 colors, coloring $(x,y)$ in color 1 if $x,y \in \mathbb{Q}$, in color 2 if $x \in \mathbb{Q}, y \notin \mathbb{Q}$ etc. – idok Feb 26 '18 at 13:24
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    See https://math.stackexchange.com/questions/2418445/infinite-hex-is-a-win-always-achievable/2418482#2418482 – Eric Wofsey Mar 01 '18 at 17:40
  • transfinite induction should be able to do it – mercio Mar 03 '18 at 21:45
  • @EricWofsey Your answer over there is excellent. Can you add it as an answer here too? I feel that this is not a duplicate question but deserves an answer which can be accepted. – M. Winter Mar 04 '18 at 16:57

1 Answers1

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Yes, this is possible using the axiom of choice. The key observation is that the image of any nonconstant path is a closed subset of $\mathbb{R}^2$ of cardinality $\mathfrak{c}$, and there are only $\mathfrak{c}$ such closed subsets. You can then construct a coloring by transfinite induction so that each such set has points of both colors.

In detail, let $(X_\alpha)_{\alpha<\mathfrak{c}}$ be an enumeration of all the closed subsets of $\mathbb{R}^2$ of cardinality $\mathfrak{c}$. We define sequences $(r_\alpha)_{\alpha<\mathfrak{c}}$ and $(b_\alpha)_{\alpha<\mathfrak{c}}$ by induction. Having defined $r_\beta$ and $b_\beta$ for all $\beta<\alpha$, define $r_\alpha$ and $b_\alpha$ to be two distinct points of $X_\alpha$ which are not equal to $r_\beta$ or $b_\beta$ for any $\beta<\alpha$. This is possible since we have chosen fewer than $\mathfrak{c}$ points so far, and $X_\alpha$ has cardinality $\mathfrak{c}$.

We thus obtain two disjoint sets $R=\{r_\alpha\}_{\alpha<\mathfrak{c}}$ and $B=\{b_\alpha\}_{\alpha<\mathfrak{c}}$ which each intersect every $X_\alpha$. Color all the points in $R$ red, and all the points in $B$ blue, and all the points that are in neither $R$ nor $B$ however you want. Then neither the red points nor the blue points contain any $X_\alpha$, and thus neither contains the image of a nonconstant path.

[This answer is adapted from my answer to Infinite 'hex': is a win always achievable?. This construction (known as "Bernstein sets") has many variants and is useful as counterexamples to many questions of this sort.]

Eric Wofsey
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