A coloring of the plane without monochromatic right triangles

TL;DR

This paper proves that under the Continuum Hypothesis, it is possible to color the plane with countably many colors without creating a monochromatic right triangle, resolving a previously claimed result.

Contribution

It provides a complete and correct proof of a coloring of the plane avoiding monochromatic right triangles under the Continuum Hypothesis, clarifying earlier claims.

Findings

Existence of a coloring with no monochromatic right triangles under CH

Complete proof of the coloring construction

Addresses and corrects previous claims by Erdős and Komjáth

Abstract

We give a full, correct proof of the following result, earlier claimed by Erd\H{o}s and Komj\'ath. If the Continuum Hypothesis holds then there is a coloring of the plane with countably many colors, with no monocolored right triangle.

Equations

$$f(x_j)\ge \max \{f(x_0),\dots ,f(x_{j-1})\}+2,$$

Full text

One2222010 Mathematics Subject Classification. Primary 03E50 Secondary 51M04, 05D10. Key words and phrases: Continuum Hypothesis, Ramsey theory of Euclidean spaces of Paul Erdős’s favorite topics consisted the applications of the Axiom of Choice to construct paradoxical sets and colorings of the Euclidean spaces. Among a large number of other questions, he raised the following: is there a coloring of the plane with countably many colors, with no monocolored right angled triangles. In [1] this was shown to be equivalent to the Continuum Hypothesis.

Recently, the senior author observed that the proof of the positive direction in [1] is incomplete. Here we give a full, correct proof.

We notice that later Schmerl in [2] gave another, more general proof, which, however, is less elementary.

Notation. Definitions. We use the notation and terminology of axiomatic set theory. The ordinals are von Neumann ordinals, $\omega_{1}$ is the least uncountable ordinal.

Theorem. (CH) There is $f:\mathbb{R}^{2}\to\omega$ with no monochromatic right angles.

Proof. Using CH, we decompose $\mathbb{R}^{2}$ and the set ${\cal E}$ of planar lines and circles into the increasing, continuous sequence of sets as $\mathbb{R}^{2}=\bigcup\{H_{\alpha}:\alpha<\omega_{1}\}$ and ${\cal E}=\bigcup\{{\cal E}_{\alpha}:\alpha<\omega_{1}\}$ such that $H_{0}={\cal E}_{0}=\emptyset$ and

(1) if $x\neq y\in H_{\alpha}$, then their connecting line and Thales circle are in ${\cal E}_{\alpha}$,

(2) if $x,y,z\in H_{\alpha}$ are not collinear, then their circuit is in ${\cal E}_{\alpha}$,

(3) if $e_{0}\neq e_{1}\in{\cal E}_{\alpha}$, then $e_{0}\cap e_{1}\subseteq H_{\alpha}$,

(4) if $x\in C\in{\cal E}_{\alpha}$, $C$ is a circle, then the antipodal point of $x$ in $C$ is in $H_{\alpha}$,

(5) if $L\in{\cal E}_{\alpha}$ is a line, $x\in H_{\alpha}\cap L$, then the line $L^{\prime}$ perpendicular to $L$ with $x\in L^{\prime}$ also contained in ${\cal E}_{\alpha}$.

This can be done by the usual Skolem-type closing arguments.

We are going to construct the coloring $f:\mathbb{R}^{2}\to\omega$ and the function $\varphi:{\cal E}\to[\omega]^{\omega}$ by transfinite recursion on $\alpha<\omega_{1}$ for $H_{\alpha+1}-H_{\alpha}$ and for ${\cal E}_{\alpha+1}-{\cal E}_{\alpha}$, satisfying the following:

(6) $f|(H_{\alpha+1}-H_{\alpha})$ is injective,

(7) if $x\in H_{\alpha+1}-H_{\alpha}$, $e\in{\cal E}_{\alpha}$, $x\in e$, then $f(x)\in\varphi(e)$,

(8) if $C\in{\cal E}$ is a circle, $i\in\varphi(C)$, $x,y\in C$, $f(x)=f(y)=i$, then $x,y$ are not antipodal,

(9) if $C\in{\cal E}$ is a circle, $i\notin\varphi(C)$, then $|f^{-1}(i)\cap C|\leq 2$,

(10) if $L\in{\cal E}$ is a line, $i\notin\varphi(L)$, then $|f^{-1}(i)\cap L|\leq 1$,

(11) if $L,L^{\prime}$ are perpendicular lines, $\{x\}=L\cap L^{\prime}$, then $f(x)\notin\varphi(L)\cap\varphi(L^{\prime})$.

Claim 1. There is no right triangle monocolored by $f$.

Proof. Assume that $x,y,z$ form a right triangle with the right angle at $y$ and $i=f(x)=f(y)=f(z)$. Let $C$ be the circle around $x,y,z$. If $i\notin\varphi(C)$, then we obtain a contradiction with (9). If $i\in\varphi(C)$, then we get a contradiction with (8), as $x,z$ are antipodal.

Claim 2. If $x\in H_{\alpha+1}-H_{\alpha}$, then there is at most one $e\in{\cal E}_{\alpha}$ such that $x\in e$.

Proof. By (3).

We add the following condition:

(12) if $x\in H_{\alpha+1}-H_{\alpha}$, $L$ is a line with $x\in L\in{\cal E}_{\alpha}$, $L^{\prime}$ is the line perpendicular to $L$ at $x$, $\{y\}=L^{\prime}\cap H_{\alpha}$, then $f(x)\neq f(y)$.

Notice that $L^{\prime}\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$ by (3) and (5), and $|L^{\prime}\cap H_{\alpha}|\leq 1$ by (1).

Assume that $f|H_{\alpha}$ and $\varphi|{\cal E}_{\alpha}$ are already constructed, we have to define $f|(H_{\alpha+1}-H_{\alpha})$ and $\varphi|({\cal E}_{\alpha+1}-{\cal E}_{\alpha})$.

Enumerate $H_{\alpha+1}-H_{\alpha}$ as $\{x_{j}:j<\omega\}$. By recursion on $j$ define $f(x_{j})$ so that

[TABLE]

$f(x_{j})$ satisfies (7), and is different from the (possible) color disqualified by (12). Clearly (6) is satisfied, and also $\omega-f[H_{\alpha+1}-H_{\alpha}]$ is infinite.

Next we define $\varphi$ on the circuits in ${\cal E}_{\alpha+1}-{\cal E}_{\alpha}$. If $C$ is such a circuit, set $A=C\cap H_{\alpha}$. By (2), we have $|A|\leq 2$.

Case 1. $|A|\leq 1$ or $f$ takes distinct values on the two elements of $A$.

In this case, set $\varphi(C)=\omega-f[A]$.

Case 2. $|A|=2$ and $f$ assumes the same value on the elements of $A$.

In this case, set $\varphi(C)=\omega$.

Finally, we define $\varphi$ on the lines in ${\cal E}_{\alpha+1}-{\cal E}_{\alpha}$. Let $L\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$ be a line. Set $B=L\cap H_{\alpha}$. Notice that by (1), $|B|\leq 1$. If $B$ is nonempty, let $y$ be its unique point.

Case 1. $B=\emptyset$ or $f(y)\notin f[L\cap(H_{\alpha+1}-H_{\alpha})]$.

Then let $\varphi(L)=\omega-f[L\cap H_{\alpha+1}]$.

Case 2. $f(y)\in f[L\cap(H_{\alpha+1}-H_{\alpha})]$.

Then set $\varphi(L)=(\omega-f[L\cap(H_{\alpha+1}-H_{\alpha}])\cup\{f(y)\}$.

Claim 3. $\varphi(e)\in[\omega]^{\omega}$ ($e\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$).

Proof. If $C\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$ is a circuit, then this is obvious in Case 2 and in Case 1 $\varphi(C)$ is $\omega$ minus at most 2 elements.

If $L\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$ is a line, the statement follows as $f[H_{\alpha+1}-H_{\alpha}]$ is a coinfinite set.

Claim 4. $\varphi$ satisfies (8).

Proof. Assume that $C\in{\cal E}_{\alpha+1}$ is a circle, $x,y\in C\cap H_{\alpha+1}$ are antipodal and $f(x)=f(y)=i$.

Case 1. $C\in{\cal E}_{\alpha}$.

In this case one of $x,y$, say $x$ must be in $H_{\alpha}$ by (6). Then, by (4), $y$ is also in $H_{\alpha}$, and we are finished by induction.

Case 2. $C\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$.

If $x,y\in H_{\alpha+1}-H_{\alpha}$, then $f(x)\neq f(y)$ by (6).

If $x,y\in H_{\alpha}$, then $C\in{\cal E}_{\alpha}$ by (1), a contradiction again.

Assume finally, that $x\in H_{\alpha}$, $y\in H_{\alpha+1}-H_{\alpha}$. If, in the definition of $\varphi(C)$, Case 1 applies, then $i\notin\varphi(C)$. We can therefore assume that Case 2 holds, $H_{\alpha}\cap C=\{x,z\}$ and $f(y)=f(z)=i$. Let $L$ be the connecting line of $x$ and $z$, $L^{\prime}$ the connecting line of $z$ and $y$.

x$$y$$z$$C$$L$$L^{\prime}

Then $L\in{\cal E}_{\alpha}$ by (1) and then $L^{\prime}\in{\cal E}_{\alpha}$ by (5). Further, by (10), $i\in\varphi(L)\cap\varphi(L^{\prime})$, which, with $f(z)=i$, contradicts (11).

Claim 5. (9) holds.

Proof. Assume that $C\in{\cal E}_{\alpha+1}$ and $i\notin\varphi(C)$. If $C\in{\cal E}_{\alpha}$, then $C\cap f^{-1}(i)$ does not contain element from $H_{\alpha+1}-H_{\alpha}$ by (7). If $C\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$ and Case 2 holds in the definition of $\varphi(C)$, then there is nothing to prove. Assume finally, that $C\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$, and Case 1 holds. Then, if $i\notin\varphi(C)$, then $C\cap f^{-1}(i)$ has at most one element in $H_{\alpha}$, at most one in $H_{\alpha+1}-H_{\alpha}$ by (6), that is at most 2.

Claim 6. (10) holds.

Proof. Assume that $L\in{\cal E}_{\alpha+1}$, $i\notin\varphi(L)$. If $L\in{\cal E}_{\alpha}$, then $L\cap f^{-1}(i)$ does not increase in $H_{\alpha+1}-H_{\alpha}$. If $L\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$ and $B=L\cap H_{\alpha}$ is empty, then $|L\cap f^{-1}(i)|\leq 1$ by (6). Otherwise, $B$ is a singleton by (1), let $y$ be its unique element. If $f(y)\notin f[L\cap(H_{\alpha+1}-H_{\alpha})]$, then again $|L\cap f^{-1}(i)|\leq 1$, otherwise we are in Case 2 of the definition of $\varphi(L)$ where we specifically added $f(y)$ to $\varphi(C)$.

Claim 7. (11) holds.

Proof. Assume that $L,L^{\prime}\in{\cal E}_{\alpha+1}$ are perpendicular, $L\cap L^{\prime}=\{x\}$, $i=f(x)\in\varphi(L)\cap\varphi(L^{\prime})$.

Case 1. $L,L^{\prime}\in{\cal E}_{\alpha}$.

In this case $x\in H_{\alpha}$ by (3), so the configuration in (11) already appers in $H_{\alpha}$, ${\cal E}_{\alpha}$.

Case 2. $L\in{\cal E}_{\alpha}$, $L^{\prime}\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$.

As $i\in\varphi(L^{\prime})$, by the definition of the latter there is $y\in L^{\prime}\cap H_{\alpha}$, $f(y)=i$. This is exactly what is ruled out at the coloring of $x$ by (12).

Case 3. $L,L^{\prime}\in{\cal E}_{\alpha+1}-{\cal E}_{\alpha}$.

Subcase 3.1. $x\in H_{\alpha}$.

As $i=f(x)\in\varphi(L)$, by the definition of $\varphi(L)$ there is $y\in L\cap(H_{\alpha+1}-H_{\alpha})$ with $f(y)=i$. Likewise, there is $z\in L^{\prime}\cap(H_{\alpha+1}-H_{\alpha})$ with $f(z)=i$. Now $y,z$ are distinct elements of $H_{\alpha+1}-H_{\alpha}$ and $f(y)=f(z)$, contradicting (6).

Subcase 3.2. $x\in H_{\alpha+1}-H_{\alpha}$.

As $i=f(x)\in\varphi(L)$, there is $y\in L\cap H_{\alpha}$, $f(y)=i$. Similarly, there is $z\in L^{\prime}\cap H_{\alpha}$, $f(z)=i$.

Let $C$ be the circuit containing $x,y,z$.

x$$y$$z$$C$$L$$L^{\prime}

As $L,L^{\prime}$ are perpendicular, $y$ and $z$ are antipodal in $C$. As in $C$ there are 3 points of color $i$, $i\in\varphi(C)$. But this contradicts to the antipodality of $y$ and $z$.

The proof of the theorem is concluded.