Proof. (IF): Let $T = (A \sqcup \bracsn{*})^{+}$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$.

For each $a \in A$, let $\pi_{a}: T \to S$ be the corresponding substitution map, and $\beta \pi_{a}: \beta T \to \beta S$ be its continuous extension. Let $p \in \beta S \setminus S$ be a minimal idempotent. By Lemma 47.5.3, there exists a minimal idempotent $q \in \beta T$ be a minimal idempotent with $pq = qp = q$ and $q \in \beta T \setminus \beta S = \beta V$.

As $\pi_{a}|_{S} = \text{Id}_{S}$, $\beta \pi_{a}|_{\beta S}= \text{Id}_{\beta S}$, so $\beta \pi_{a}(p) = p$. Since $q$ is idempotent and $\beta \pi_{a}$ is a semigroup homomorphism, $\beta \pi_{a}(q) \in \beta S$ is also idempotent. Further,

and

so $\beta \pi_{a}(q)p = p\beta \pi_{a}(q) = \beta \pi_{a}(q)$. In particular, $\beta \pi_{a}(q) = \beta \pi_{a}(q)p \in (\beta S)p$, where $(\beta S)p$ is a minimal left ideal in $\beta S$. By minimality, $(\beta S)\beta \pi_{a}(q) = (\beta S)p$, and there exists $u \in \beta S$ with $p = u\beta \pi_{a}(q)$. As such,

Let $S = \bigsqcup_{j = 1}^{r} C_{r}$ be a finite colouring of $S$, then there exists a unique $1 \le i \le r$ such that $C_{i} \in p$. Since $\beta \pi_{a}(q) = p$ for all $a \in A$, $\bigcap_{a \in A}\pi_{a}^{-1}(C_{i}) \in q$. As $q \in \beta T \setminus \beta S = \beta V$, $V \in q$, and $V \cap \bigcap_{a \in A}\pi_{a}^{-1}(C_{i}) \ne \emptyset$. Therefore any element of $V \cap \bigcap_{a \in A}\pi_{a}^{-1}(C_{i})$ generates a combinatorial line, contained in $C_{i}$.

(F): Let $r \ge 1$ and suppose for contradiction that for every $N \in \natp$, there exists an $r$-colouring $c_{N}: A^{N} \to \bracsn{1, \cdots, r}$ such that $c_{N}$ contains no monochromatic combinatorial line. Define

then $c$ is a $r$-colouring of $S$ containing no monochromatic combinatorial lines, which contradicts (IF).$\square$