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?