Extension of the Paris-Harrington principle Let $[m,n]$ denote the set $\{m,m+1, ... ,n-1,n\}$. $X \to (k)^n_c$ means that whenever $f: [X]^n \to c$ there is a subset $H \subset X$ with cardinality $k$ such that $f$ is constant on $[H]^n$ (The set $H$ here is often described as being homogeneous for $f$.) The Paris-Harrington principle (PH) stipulates further that the homogeneous set should be 'relatively large'. Let $X \to_* (k)^n_c$ mean that whenever $f: [X]^n \to c$ there is $H \subset X$ that is homogeneous for $f$ and $\text{card}(H) \geq \text{max}(k, \text{min}(H))$. PH states that $\forall n,c,k \in \mathbb{N} \; \exists m \in \mathbb{N} \; (m \to_* (k)^n_c)$.
Is the following true? And how do you prove it from PH?
$$ \forall n,c,k,a \in \mathbb{N} \; \exists m \in \mathbb{N} \; ([a, m] \to_* (k)^n_c) $$
 A: The principle
$$\forall n,c,k,a\in\omega\,\exists m\in\omega\,\big([a,m]\to_*(k)_c^n\big)\tag{1}$$
is indeed true; it can be proved by essentially the same proof that Paris and Harrington gave for their principle.
Fix $n,c,k$, and $a$, and suppose that no such $m$ exists. Call $\xi$ a counterexample for $m$ if $\xi$ is a $c$-coloring of $[a,m]^n$ with no relatively large homogeneous set of cardinality at least $k$. Let $T_m$ be the set of all counterexamples for $m$, and let $T=\bigcup_{m\ge a}T_m$; then $\langle T,\subseteq\rangle$ is a finitely branching tree of height $\omega$, so by König’s lemma there is a branch $B$ through it. Let $\xi=\bigcup B$; $\xi$ is a $c$-coloring of $[\omega\setminus a]^n$ such that for each $m\ge a$, $\xi_m=\xi\upharpoonright[a,m]^n$ is a counterexample for $m$.
The infinite Ramsey theorem ensures that there is an infinite $H\subseteq\omega\setminus a$ such that $[H]^n$ is homogeneous for $\xi$. Clearly we can choose $m$ large enough that
$$\big|H\cap[a,m]\big|\ge\max\{k,\min H\}\,,$$
contradicting the assumption that $\xi_m$ is a counterexample for $m$ and establishing $(1)$.
I do not yet see how to prove it directly from PH, however.
