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Every point of three-dimensional space is colored red, green, or blue. Prove that one of the colors attains all distances, meaning that any positive real number represents the distance between two points of this color.

My proof: Suppose by contradiction that $\exists \delta>0$ such that for any $x,y\in \mathbb{R}^3$ with $d(x,y)=\delta$ points $x$ and $y$ have different colors. Let's consider the tetrahedron in $\mathbb{R}^3$ with base $A_1A_2A_3$ and upper vertex $A_4$. Suppose $A_1$ is colored into red,$A_2$ is colored into green then $A_3$ is colored into blue. Then $A_4$ should be colored into one of the colors blue, red, or green but due to $d(A_4,A_1)=d(A_4,A_2)=d(A_4,A_3)=\delta$ the coloring of point $A_4$ to red,blue, or green is impossible.

EDIT: Let's prove that RED attains all distances.Suppose by contradiction that $\exists \delta>0$ such that for any $x,y\in \mathbb{R}^3$ with $d(x,y)=\delta$ points $x$ and $y$ not both RED. Let's consider the regular tetrahedron in $\mathbb{R}^3$ with base $A_1A_2A_3$ and upper vertex $A_4$. WLOG suppose $A_1$ is colored into red,$\ A_2$ is colored into green then $A_3$ is colored into blue. Then $A_4$ should be colored into one of the colors blue, red, or green but due to $d(A_4,A_1)=d(A_4,A_2)=d(A_4,A_3)=\delta$ but the coloring of point $A_4$ to red,blue, or green is impossible. Here we get contradiction.

Right?

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  • $\begingroup$ Your edit is really messy to be honest. $\endgroup$ – Fimpellizieri Nov 18 '17 at 6:33
  • $\begingroup$ @Fimpellizieri, Is it incorrect? $\endgroup$ – ZFR Nov 18 '17 at 6:42
  • $\begingroup$ Your proof still does not work. You cannot prove red attains all distances; there might not even be any red points. A proper proof by contradiction would start like this: “assume there exist numbers $r,b,g>0$ so that no two red points are $r$ apart, no two blue points are $b$ apart, and no two green points are $g$ apart.” $\endgroup$ – Mike Earnest Nov 18 '17 at 16:22
  • $\begingroup$ @MikeEarnest, could you explain in detail why my proof (in EDIT) does not work? What kind of flaws it contain? I would like to know it. Would be grateful. In my opinion it's OK $\endgroup$ – ZFR Nov 18 '17 at 16:34
  • $\begingroup$ It is not a contradiction for $A_4$ to be blue or green; you've only supposed that no two RED points have a distance of $\delta$. Also, you cannot say that WLOG $A_1$ is red, since you have singled out red as a special color. $\endgroup$ – Mike Earnest Nov 18 '17 at 16:48
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You have proven that for any distance $\delta$, we can find two points at that distance whose colors match. What you haven't proven (and what is requested) is that we can always find two points at that distance that are both a particular color (red, for example). Your construction leaves open the possibility that for (say) $\delta = 1$ all pairs thus constructed are blue, while for $\delta = 2$ all pairs thus constructed are red.

EDIT: I'm not entirely sure what the logic is in your revised proof; the concluding sentences are rather unclear to me. However, it appears to assume that the vertices of any tetrahedron include all three colors, but this is not necessarily the case.

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  • $\begingroup$ Yes, you are right. I did not pay attention to this serious detail. $\endgroup$ – ZFR Nov 18 '17 at 6:08
  • $\begingroup$ I have edited my post. Please take a look. $\endgroup$ – ZFR Nov 18 '17 at 6:20
  • $\begingroup$ @RFZ: I still don't think it works; I've edited my answer to explain why. $\endgroup$ – Michael Seifert Nov 18 '17 at 20:23
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You're on the right track, but your proof could do with a bit of tidying to make it clearer.

Suppose for contradiction that $\exists \delta>0$ such that for any $x,y\in \mathbb{R}^3$ with $d(x,y)=\delta$ points $x$ and $y$ have different colors.

Consider the regular tetrahedron in $\mathbb{R}^3$ with all side lengths equal to $\delta$, and label the base $A_1A_2A_3$ and upper vertex $A_4$. Without loss of generality, take $A_1$ coloured red. Then $A_2$ must be differently coloured, say it is green. Then $A_3$ must be coloured differently to red and green, so say it is blue. But now, $A_4$ cannot be coloured red, green or blue, and we reach our desired contradiction.

The two important changes I've made are to specify the tetrahedron has side length $\delta$, and explain why we took $A_1$ red, $A_2$ green and so on.

EDIT: Michael Seifert points out more important issues with your proof which I missed.

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  • $\begingroup$ Yes, you are right. I have proven that any distance is attained by points whose colors match. But i need to prove that this color is certain, for example, red. I have edited my post above. Please take a look. $\endgroup$ – ZFR Nov 18 '17 at 6:23
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Suppose there is no color which attains all distances. This means there exists a distance $r$ not attained by any two red points, a distance $b$ not attained by any two blue points, and same for $g$ with green points. WLOG $r\ge b\ge g$.

Let $x$ be any red point*. Let $S_x$ be the (surface of the) sphere of radius $r$ centered at $x$. Every point in $S_x$ must be blue or green.

Let $y$ be any blue point in $S_x$**. Let $S_y$ be the sphere of radius $b$ centered at $y$. Since $b\le r$, the intersection of $S_x$ and $S_y$ is a circle. Every point in that circle must be green.

After drawing a picture, you should be able to convince yourself that the diameter of this circle is at least $b$; since $b\ge g$, there will exist two points on the circle whose distance is $g$, which contradicts the supposed fact that no two green points are at distance $g$.

*Here we are assuming that there exists a red point. What if there are no red points? This is left as an exercise to the reader.

**What do you do if no blue point exists?

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  • $\begingroup$ This appears to nicely generalize to $n$-colorings of $\mathbb{R}^n$, although I haven't really checked. $\endgroup$ – Fimpellizieri Nov 18 '17 at 22:41

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