= Halting-stage colouring of triples
Let $K_s$ be an increasing uniformly computable finite approximation to the <diagonal halting set>. For $a<b<c$, colour the triple $0$ if $K_b\restriction a=K_c\restriction a$, and $1$ otherwise. No infinite <homogeneous set for a colouring> can have colour $1$, since the finitely many bits below its first element eventually stabilize. Every infinite homogeneous set $H$ of colour $0$ computes the <diagonal halting set>: choose $n<a<b$ in $H$ and read membership of $n$ in $K_b$. Homogeneity and unboundedness make this answer final. Thus the colouring has no infinite computable homogeneous set.
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