Proof.

• $\Rightarrow$ If $X$ is Euclidean Ramsey then take $S$ finite in ${ℝ}^n$ (for $k$-colours).
• $\Leftarrow$ We know that for all $k$-colourings of ${ℝ}^n$, there exists a monochromatic copy of $X$ (by compactness). Suppose not. Therefore, for any set $S$ finite in ${ℝ}^n$, there exists a bad colouring (i.e. not a copy of $X$ monochromatic). Space of all $k$-colourings is ${[k]}^{{ℝ}^n}$, which is compact by Tychonoff. Let

  $$C_{X^{\prime }}=\{\text{colourings that do not make }{X}^{\prime }\text{ monochromatic}\}.$$

  This is a closed set. Let ${\{C_{X^{\prime }}\}}_{X^{\prime }\text{ a copy of }X}$. It has the finite intersection property, because any finite $S$ has a bad $k$-colouring. Hence the intersection of all $C_{X^{\prime }}$ is non-empty. Hence there exists a colouring of ${ℝ}^m$ with no monochromatic $X$, contradiction. □

Proof. Let $k$ be a colouring of $S\times T$, where $S$ is $k$-Ramsey for $X$ and $T$ is $k^{|S|}$-Ramsey for $Y$.

We $k^{|S|}$-colour $T$ as follows:

By choice of $T$, there exists a monochromatic $Y^{\prime }$ (copy of $Y$) with respect to $c^{\prime }$, i.e. $c(s,y)=c(s,y^{\prime })$ for all $y,y^{\prime }\in Y$ and any $s\in S$.

Now $k$-colour $S$ via $c^{″}(s)=c^{″}(s,y)$ for some $y\in Y^{\prime }$ (which is well-defined by the above). By the choice of $X$, there exists monochromatic $X^{\prime }$ with respect to $c^{″}$, and hence $X^{\prime }\times Y^{\prime }$ is monochromatic with respect to $c$. □

Proof. Recall in ${ℝ}^n$ we have

Every copy of $\{0,1,2\}$ is $\{x-y,x,x+y\}$ with $\parallel y{\parallel }_2=1$ (in any ${ℝ}^n$). We have

If we can find a colouring of ${ℝ}_{\ge 0}$ such that there does not exist a monochromatic solution to $a+b=2c+2$. Then use $c(x)\to \varphi (\parallel x{\parallel }_2^2)$.

We $4$-colour ${ℝ}_{\ge 0}$ by $\varphi (x)=\lfloor x\rfloor (mod4)$.

Suppose $a,b,c$ all have colour $n\in \{0,1,2,3\}$. Then $a+b=c+2$ implies

where $M_4$ is a multiple of $4$. This is impossible as $-2<\{a\}+\{b\}-2\{c\}<2$. □

Proof.

• $\Rightarrow$ $x_1,\dots ,x_d$ are not spherical. The first two conditions say that $x_1,\dots ,x_d$ is not a simplex: so there exists $c_1,\dots ,c_{d-1}$ (not all $0$) such that ${\sum }_i{c}_i(x_i-x_d)=0$ or ${\sum }_{i=1}^{N-1}{c}_i{x}_i+(-c_1-c_2-\cdots -c_{d-1})x_d=0$.

  First we note that (1)–(3) are invariant under translation by $v\in {ℝ}^n$:

  • ${\sum }_i{c}_i(x_i+v)=0$ since ${\sum }_i{c}_i=0$

  • ${\sum }_i{c}_i\parallel x_i+v{\parallel }_2^2={\sum }_i{c}_i\parallel x_i{\parallel }_2^2+2c_i(x_i,v)+c_i\parallel v{\parallel }_2^2={\sum }_i{c}_i\parallel x_i{\parallel }_2^2+2\left({\sum }_i{c}_i{x}_i,v\right)+\parallel v{\parallel }_2^2{\sum }_i{c}_i={\sum }_i{c}_i\parallel x_i{\parallel }_2^2$

  Let us look at a minimal subset of $x_1,\dots ,x_d$ that is not spherical. If we can show this for without loss of generality $x_1,\dots ,x_k$ ($k\le d$) then take $c_i=0$ for all $i\in [k+1,d]$.

  Since $\{x_1,\dots ,x_k\}$ are not spherical, it is not a simplex. Hence there exists $c_i$ such that ${\sum }_i{c}_i(x_i-x_k)=0$. Then

  $$c_1{x}_1+c_2{x}_2+\cdots +c_{k-1}{x}_{k-1}+(-c_1-c_2-\cdots -c_{k-1})x_k=0.$$

  Without loss of generality $c_1\ne 0$ (because the conditions are invariant under reordering the $c_i$’s).

  Consider $\{x_2,\dots ,x_k\}$. This is spherical by minimality. Suppose the sphere radius is $r$, and centred at $y$. By the above, translate the set such that $\{x_2,\dots ,x_k\}$ is centred at $0$. This is fine because the same $c_i$’s work after translations ((1)–(3) are totally invariant under translations). Then

  $$\sum _{i=1}^k{c}_i\parallel x_i{\parallel }_2^2=c_1\parallel x_1{\parallel }_2^2+r^2\left(\sum _{i=2}^k{c}_i\right)\ne 0$$

  as $\parallel x_1\parallel \ne r$.

• $\Leftarrow$ Suppose there exists $c_i$ as in the statement, and assume $(x_1,\dots ,x_d)$ are spherical, centred at $v$, radius $r$.

  Translate the set so that they are centred at the origin (this preserves all conditions and does not vhange the value of ${\sum }_I{c}_i\parallel x_i{\parallel }^2\ne 0$).

  Let $\parallel x_i{\parallel }^2=r^2$. Then

  $$\sum _i{c}_i\parallel x_i{\parallel }^2=r^2\sum _i{c}_i=0$$

  so $(x_1,\dots ,x_d)$ are not spherical. □

Proof. Suppose $X$ is not spherical. Then there exists $c_i$ (not all $0$) such that ${\sum }_i\parallel x_i{\parallel }^2=b$ and ${\sum }_i{c}_i=0$.

Also true for any copy of $X^{\prime }$.

Going to split $[0,1)$ into $[0,\delta ),[\delta ,2\delta ),\dots$ for small $\delta$. Then colour depending on where $c_i\parallel x{\parallel }^2$ lies.

Let’s prove that there is a colouring of $ℝ$ such that ${\sum }_{i=1}^n{c}_i{y}_i=b$ does not have a monochromatic solution. Let ${\sum }_{i=1}^{n-1}{c}_i(y_i-y_n)=b$. By rescaling, we may assume $b=\frac{1}{2}$. Now we split $[0,1)$ into intervals $[0,\delta ),[\delta ,2\delta ),\dots$ where $\delta$ is very small. Let

A ${\left(\lfloor \frac{1}{\delta }\rfloor \right)}^{n-1}$-colouring.

Assume $y_1,\dots ,y_{n-1}$ monochromatic under $c$ and such that $\sum c_i(y_i-y_n)=\frac{1}{2}$.

Hence the sum is within $(n-1)\delta$ of an integer, so not $\frac{1}{2}$, if $\delta$ small enough. □

Proof. (Yawn, product argument…)

We will by induction on $n$ (and all $k$ at once).

$n=1$ is just the assumption.

Suppose true for $n-1$. Fix $k$. Let $S$ be a finite set such that whenever $S$ is $k^{|X|}$ coloured, there exists an $A$-invariant copy of $X^{n-1}$.

$T$ a finite set such that whenever $T$ is $k^{|S|}$-coloured, there exists a copy of $X$ with $A$ monochromatic.

Claim: $S\times T$ works for $X^n$.

By definition of $T$, if we look at $c^{\prime }(s,t)=(c(s_1,t),c(s_2,t),\dots ,c(s_{|S|},t))$, a $k^{|S|}$-colouring. Thus there exists a copy of $X$ with $A$ monochromatic.

This induces a colouring of $S$ as follows: $c^{″}(s)=(c(s,a),c(s,x_1),\dots ,c(s,x_{|X|-|A|}))$ for some $a\in A$ (note $c(s,a)$ is well-defined). This is a $k^{|X|-|A|+1}$-colouring.

Therefore by the choice of $n$, there exists a copy of $X^{n-1}$ that is $A$-invariant.

Then we are done as this copy of $X$ in $T$ is $A$-invariant. □

Proof. $X=\{1,2,\dots ,m\}$, where the numbers are the names of points.

We find a copy of $\sqrt{m}X$ of the form

We will show by induction on $r$ and all $k$ at once that we can find a copy of $X$ with $\{1,2,\dots ,r\}$ monochromatic.

$r=1$ is trivial as it is just a point.

$r=2$ is 2 points at a specified distance (which we showed is Ramsey).

Assume true for $r$ and all $k$. By Our product argument, there exists $S$ and $N$ such that we have a $X^N$ (copy) on which the colouring is $\{1,2,\dots ,r\}$-invariant on $X^N$ (for any $N$). We will choose $N$ to be as big as we want.

The clever bit:

We will colour $(m-1)$ sets in $[N]$ say $a_1<a_2<\cdots <a_{m-1}$ as follows.

$c^{\prime }(\{a_1,\dots ,a_{m-1}\})=(c(w_1),c(w_2),\dots ,c(w_r))$ is a $k^r$ colouring of ${[N]}^{(m-1)}$.

As $N$ can be taken as big as needed, by Ramsey there exists a $m$-monochromatic set. By relabeling, we may assume that this set is $\{1,2,\dots ,m\}$ coordinates.

Now we look at the following:

with this we note that the colour of $y_i$ is the same as the colour of $x_{i+1}$.

Now look at: $(1,2,\dots ,m)$, $2,3,\dots ,m,1)$, …, $(r+1,\dots ,r,r-1)$. monochromatic copy of $\{1,2,\dots ,r+1\}$. They all have the same colour (ignoring this). □