[proofplan] The result is a finite analogue of the [Infinite Ramsey Theorem for r-Sets](/theorems/2049), and the proof parallels that construction but tracks sizes. We use a double induction: outer on $r$ and, within each $r$, inner on $s + t$. The base $r = 1$ reduces to the pigeonhole principle; the base of the inner induction (with $s \leq r$ or $t \leq r$) holds trivially. The inductive step picks any vertex $x$, looks at the induced $(r-1)$-colouring $c_x(F) := c(\{x\} \cup F)$ on the remaining vertices, and applies the $(r-1)$-case to find either a red "candidate" block of size $R^{(r)}(s-1,t)$ or a blue block of size $R^{(r)}(s,t-1)$. Inside that block we apply the $r$-case at lower $s+t$ (available by inner induction) and combine with $x$ to produce the desired monochromatic $r$-subset. [/proofplan] [step:Set up the double induction and dispose of the boundary cases] We show by simultaneous induction that $R^{(r)}(s,t)$ is finite for all $r, s, t \geq 1$, together with the recursive bound. The outer variable is $r$; the inner variable is $s + t$. *Case $r = 1$.* A 2-colouring of $\{1, \ldots, n\}^{(1)}$ is a function $c : \{1, \ldots, n\} \to \{\text{red}, \text{blue}\}$. By the pigeonhole principle, if $n \geq (s - 1) + (t - 1) + 1 = s + t - 1$, then either at least $s$ elements receive colour red or at least $t$ elements receive colour blue. Hence $R^{(1)}(s, t) \leq s + t - 1 < \infty$. *Boundary cases for $r \geq 2$.* If $s \leq r$, then any $s$-element subset of $\{1, \ldots, n\}$ with $n \geq s$ contains no $r$-subset unless $s = r$, in which case the $s$-subset itself is a single $r$-subset and is trivially monochromatic. More precisely: - If $s < r$, then every $s$-subset has $\binom{s}{r} = 0$ many $r$-subsets, so the condition "all $r$-subsets red" is vacuous; any $s$ vertices form a "red $s$-set." Hence $R^{(r)}(s, t) \leq s$. - If $s = r$, then an $s$-subset has exactly one $r$-subset, namely itself, which is either red or blue. If $n \geq s$, pick any $r$-subset; if it is red, we have our red $s$-set, and if it is blue, we need a blue $t$-set, handled symmetrically by $t \leq r$ or by $n \geq t$. In all cases $R^{(r)}(s, t) \leq \max(s, t)$ when $\min(s, t) \leq r$, so the boundary is finite. *Inductive hypotheses.* Fix $r \geq 2$ and suppose $R^{(r')}(s', t')$ is finite for all $r' < r$ and all $s', t' \geq 1$ (outer hypothesis). Within the fixed $r$, fix $s, t \geq r + 1$ and suppose $R^{(r)}(s', t')$ is finite for all $s', t' \geq 1$ with $s' + t' < s + t$ (inner hypothesis). [/step] [step:Define the candidate $N$ and fix a 2-colouring of $\{1, \ldots, N\}^{(r)}$] Set \begin{align*} N := R^{(r-1)}\!\left(R^{(r)}(s-1, t),\ R^{(r)}(s, t-1)\right) + 1. \end{align*} Both arguments of $R^{(r-1)}$ are finite by the inner hypothesis ($R^{(r)}(s-1, t)$ and $R^{(r)}(s, t-1)$ use strictly smaller $s + t$), and $R^{(r-1)}$ is finite by the outer hypothesis, so $N < \infty$. Let $n \geq N$ and let $c : \{1, \ldots, n\}^{(r)} \to \{\text{red}, \text{blue}\}$ be any 2-colouring. We show $c$ admits either a red $s$-subset (all of whose $r$-subsets are red) or a blue $t$-subset (all of whose $r$-subsets are blue). [/step] [step:Reduce to an $(r-1)$-colouring by fixing a vertex $x$] Pick $x := 1 \in \{1, \ldots, n\}$. Consider the induced colouring \begin{align*} c_x : (\{1, \ldots, n\} \setminus \{x\})^{(r-1)} &\to \{\text{red}, \text{blue}\} \\ F &\mapsto c(\{x\} \cup F). \end{align*} This is a 2-colouring of the $(r-1)$-subsets of a set of size $n - 1 \geq N - 1 = R^{(r-1)}(R^{(r)}(s-1, t),\ R^{(r)}(s, t-1))$. By definition of $R^{(r-1)}$, at least one of the following holds: (i) there exists a set $S_1 \subseteq \{1, \ldots, n\} \setminus \{x\}$ with $|S_1| = R^{(r)}(s-1, t)$ such that $S_1^{(r-1)}$ is entirely red under $c_x$; (ii) there exists a set $S_2 \subseteq \{1, \ldots, n\} \setminus \{x\}$ with $|S_2| = R^{(r)}(s, t-1)$ such that $S_2^{(r-1)}$ is entirely blue under $c_x$. [guided] We want to reduce the problem for $r$-subsets to a problem for $(r-1)$-subsets so we can invoke the outer inductive hypothesis. The key observation: if we *fix* one vertex $x$, then an $r$-subset containing $x$ is determined by choosing an $(r-1)$-subset $F$ from the remaining vertices, and its $c$-colour is $c(\{x\} \cup F)$ — a function of $F$ alone. This induced map \begin{align*} c_x : (\{1, \ldots, n\} \setminus \{x\})^{(r-1)} &\to \{\text{red}, \text{blue}\} \\ F &\mapsto c(\{x\} \cup F) \end{align*} is a 2-colouring of the $(r-1)$-subsets of a set of size $n - 1$. We chose $N$ so that $n - 1 \geq R^{(r-1)}(R^{(r)}(s-1, t), R^{(r)}(s, t-1))$ — large enough that the $(r-1)$-dimensional Ramsey function, with cleverly chosen colour targets, guarantees either a large red or a large blue set under $c_x$. Why those particular colour targets? Because after finding a large $c_x$-monochromatic set $S$, we want to apply the $r$-Ramsey theorem *again* (at smaller $s + t$) inside $S$ under the original colouring $c$. For the red case we want $|S_1|$ at least $R^{(r)}(s-1, t)$, and for the blue case $|S_2|$ at least $R^{(r)}(s, t-1)$ — the sizes that let the inner induction produce either an $(s-1)$-red set to combine with $x$, or a $t$-blue set outright. The symmetric pairing explains the arguments of the outer $R^{(r-1)}$. By definition of $R^{(r-1)}$ with these two target sizes, at least one of the following occurs: (i) there exists $S_1 \subseteq \{1, \ldots, n\} \setminus \{x\}$ with $|S_1| = R^{(r)}(s-1, t)$ and all $(r-1)$-subsets of $S_1$ red under $c_x$; (ii) there exists $S_2 \subseteq \{1, \ldots, n\} \setminus \{x\}$ with $|S_2| = R^{(r)}(s, t-1)$ and all $(r-1)$-subsets of $S_2$ blue under $c_x$. These two cases are symmetric; we analyse (i) and then indicate the change for (ii). [/guided] [/step] [step:In the red case, apply the $r$-Ramsey theorem to $S_1$ at smaller $s + t$] Assume case (i) holds. Restrict the original colouring $c$ to $r$-subsets of $S_1$. Since $|S_1| = R^{(r)}(s-1, t)$ and $(s-1) + t = s + t - 1 < s + t$, the inner inductive hypothesis applies to $S_1$. Hence either: (a) there exists $A \subseteq S_1$ with $|A| = s - 1$ and $A^{(r)}$ entirely red under $c$; or (b) there exists $T \subseteq S_1$ with $|T| = t$ and $T^{(r)}$ entirely blue under $c$. In subcase (b), $T$ is already the desired blue $t$-subset, and we are done. In subcase (a), we claim that $A \cup \{x\}$ is an $s$-element set with $(A \cup \{x\})^{(r)}$ entirely red. Since $x \notin S_1 \supseteq A$, $|A \cup \{x\}| = s$. Let $R \in (A \cup \{x\})^{(r)}$. Two cases: - *$x \notin R$.* Then $R \subseteq A$, so $c(R)$ is red by property of $A$. - *$x \in R$.* Then $R = \{x\} \cup F$ with $F \in A^{(r-1)} \subseteq S_1^{(r-1)}$. By case (i), $c_x(F) = c(\{x\} \cup F) = c(R)$ is red. In either case $c(R)$ is red, so $(A \cup \{x\})^{(r)}$ is monochromatic red with $|A \cup \{x\}| = s$, as required. [/step] [step:Handle the blue case by symmetry and conclude] Assume case (ii). By the same argument with the colours red and blue swapped and $(s-1, t)$ replaced by $(s, t-1)$: since $|S_2| = R^{(r)}(s, t-1)$ and $s + (t-1) < s + t$, the inner hypothesis applied to $S_2$ yields either a red $s$-subset of $S_2$ (done immediately) or a blue subset $B \subseteq S_2$ with $|B| = t - 1$. In the latter case, $B \cup \{x\}$ has size $t$, and every $r$-subset of $B \cup \{x\}$ is blue: either it lies in $B$ (blue by choice of $B$) or it equals $\{x\} \cup F$ with $F \in B^{(r-1)} \subseteq S_2^{(r-1)}$, hence blue by case (ii). Thus in every case $\{1, \ldots, n\}$ contains either a red $s$-subset or a blue $t$-subset. This holds for every $n \geq N$, so $R^{(r)}(s, t) \leq N$, completing the inner induction step and the proof of the recurrence \begin{align*} R^{(r)}(s, t) \leq R^{(r-1)}\!\left(R^{(r)}(s-1, t),\ R^{(r)}(s, t-1)\right) + 1. \end{align*} Finiteness follows by induction from finiteness of the right-hand side. [/step]