(If $\mathbb{N}$ is finitely coloured, there exists ${(x_n)}_{n=1}^{\infty }$ such that $\mathrm{FS}(x_1,x_2,\dots )$ is monochromatic).

Let $A_1,A_2,\dots ,A_k$ be the colour classes ($A_1\bigsqcup A_2\bigsqcup \cdots \bigsqcup A_k=\mathbb{N}$) and $U$ an idempotent ultrafilter.

Claim: $A_i\in U$ for some $i$ (this is because ultrafilters are prime: whenever we have a finite union lying in the ultrafilter we have at least one of the components lying in the ultrafilter, else have $A_i^c\in U$ for each $i$, hence $A_1^c\cap \cdots \cap A_k^c\in U$, but this is ${(A_1\cup \cdots \cup A_k)}^c$, contradicting the fact that $A_1\cup \cdots \cup A_k\in U$).

Let $A=A_i\in U$. Therefore we have ${\forall }_{U}y,y\in A$. Then:

• (1) ${\forall }_{U}x,{\forall }_{U}y,y\in A$
• (2) ${\forall }_{U}x,{\forall }_{U}y,x\in A$
• (3) $A\in U+U=U$ gives that ${\forall }_{U}x,{\forall }_{U}y,x+y\in A$.

Then (1) and (2) and (3) give:

$${\forall }_{U}x,{\forall }_{U}y,\mathrm{FS}(x,y)\subseteq A.$$

Now fix $x_1\in A$ such that

$${\forall }_{U}y,\mathrm{FS}(x_1,y)\subseteq A.$$

Assume we have found $x_1,\dots ,x_n$ such that

$$\begin{array}{llll}\hfill {\forall }_{U}y,\mathrm{FS}(x_1,\dots ,x_n,y)& \subseteq A& \hfill & \\ \hfill \underset{=:B}{\underbrace{\{y:\mathrm{FS}(x_1,\dots ,x_n,y)\subseteq A\}}}& \in U=U+U& \hfill & \end{array}$$

Then have ${\forall }_{U}x,{\forall }_{U}y,x+y\in B$.

• (1) ${\forall }_{U}x,{\forall }_{U}y,\mathrm{FS}(x_1,\dots ,x_n,x+y)\subseteq A$
• (2) ${\forall }_{U}x,{\forall }_{U}y,\mathrm{FS}(x_1,\dots ,x_n,x)\subseteq A$
• (3) ${\forall }_{U}x,{\forall }_{U}y,\mathrm{FS}(x_1,\dots ,x_n,y)\subseteq A$.

Then (1), (2) and (3) give:

$${\forall }_{U}x,{\forall }_{U}y,\mathrm{FS}(x_1,\dots ,x_n,x,y)\subseteq A.$$

Thus fix $x_{n+1}\gg x_n$. Have ${\forall }_{U}y,\mathrm{FS}(x_1,\dots ,x_n,x_{n+1},y)\subseteq A$. Then done by induction.