Let me state the notations first. It is pretty long and contains complicated notations.
$n,r, k, s_i$ are positive integers and $s_1, \cdots, s_r \geq k$. $[n]$ denotes the set $\{1,2,\cdots,n\}$. For set $A$ and an integer $m$, ${A \choose m}$ denotes set of all $m$-subsets of $A$.
Let $\mathcal{C}(n,r,k) = \left\{ c : {[n] \choose k} \to [r] \right\}$. This means all possible colorings with $r$ colors for $k$-subsets of $[n]$. For each $c \in \mathcal{C}(n,r,k)$, define a predicate $\alpha(c,s_1, \cdots, s_r)$ to be true iff $\exists i \in [r]$, $\exists A \in {[n] \choose s_i}$ s.t. $c(B)=i$ for all $B \in {A \choose k}$.
Now let $R_r(k;s_1, \cdots, s_r) = \min\{n>0:\alpha(c,s_1, \cdots, s_r) \text{ is true for all } c \in \mathcal{C}(n,r,k)\}$.
This is the smallest integer $n$ with the property that, for arbitrary coloring of $k$-subset of $[n]$, then for some $i \in [r]$, there is an $s_i$-subset of $[n]$, all of whose $k$-subsets have color $i$.
Let $R_r(k;s) = R_r(k;s_1, \cdots, s_r)$ where $s_1 = \cdots = s_r =s$.
Now begins the problem. The objective is to show the finiteness of $R_r(k;s_1, \cdots, s_r)$. Show the following:
(a) $R_2(k;s,t) \leq R_2(k-1;R_2(k;s-1,t), R_2(k;s,t-1))+1$
(b) $R_{r+1}(k;s) \leq R_r(k; R_2(k;s))$
(c) $R_r(k;s_1, \cdots, s_r)$ is finite for all possible choices of $k,r,s_i$'s.
For (a), hint suggests to use induction on the order on $\{k,s,t\}$ defined by:
$(k,s,t) \prec (k', s', t')$ if $k < k'$, and if $k=k'$, then $(k,s,t) \prec (k', s', t')$ when $s+t \leq s'+t'$.
This is way too different from the problems I encountered before. How can I use induction on arbitrary order? And how can I exploit the induction to solve this problem?
Thanks in advance for any form of help, hint, or solution.
Let $C_a(k, s, t)$ be the indicator variable of $(a)$ w.r.t $k$, $s$, and $t$. Specifically, $C_a(k, s, t)$ is True if and only if (a) holds w.r.t $k$, $s$, and $t$.
Starting point
When $k = 1$ or $s = k$ or $t = k$, it is trivial. The first nontrivial case is $(k, s, t) = (2, 3, 3)$. Claim: $C_a(2, 3, 3)$ is True.
To see this, when $(k, s, t) = (2, 3, 3)$, $$R_2(2; 3, 2) = R_2(2; 2, 3) = 3$$ and then $$RHS = R_2(1; 3, 3) = 5$$ while $$LHS = R_2(2; 3, 3) = 6.$$
Proceeding
We shall show that given $k$, $s$, and $t$, if $C_a(k', s', t') = True$ for all $(k', s', t')$ such that $(k', s', t') \prec (k, s, t)$ and $(k', s', t') \neq (k, s, t)$, then $C_a(k, s, t) = True$.
Choose arbitrarily a RHS-set $S_R \in \binom{[n]}{RHS}$, and arbitrarily choose $s_R \in S_R$, then choose arbitrarily a 2-coloring $c_R$ on $\binom{S_R}{k}$, we shall show that $c_R$ gives a monochromatic color-$1$ $s$-set or a monochromatic color-$2$ $t$-set.
The induction hypothesis is used so that RHS is finite and both $R_2(k;s-1,t)$ and $R_2(k;s,t-1)$ are finite.
From $c_R$, we define a 2-coloring $c'_R$ on $S'_R := \binom{S_R \setminus \{s_R\}}{k-1}$ with $|S'_R| = |S_R| - 1 = R_2(k-1;R_2(k;s-1,t), R_2(k;s,t-1))$ by the following: for each $S' \in \binom{S'_R}{k-1}$ $(k-1)$-set, we let $$c'_R(S') = c_R(S' \cup \{s_R\}).$$ By the definition, $c'_R$ must give a monochromatic color-$1$ $R_2(k;s-1,t)$-set or a monochromatic color-$2$ $R_2(k;s,t-1)$-set. WLOG, we assume that there is a monochromatic color-$1$ $R_2(k;s-1,t)$-set, and let $X$ be such a $R_2(k;s-1,t)$-set.
For this $X$, under $c_R$, there should be a monochromatic color-$1$ $(s-1)$-set or a monochromatic color-$2$ $t$-set. If there is a monochromatic color-$2$ $t$-set then LHS is satisfied. If there is a monochromatic color-$1$ $(s-1)$-set $Y$, then consider $Z := Y \cup \{s_R\}$, recall that we let $$c'_R(S') = c_R(S' \cup \{s_R\}),$$ which implies that $Z$ is a monochromatic color-$1$ $s$-set, completing the proof.
I think this is Ramsey number.