[proofplan] We proceed by induction on $r$. The base case $r = 2$ is the [Infinite Ramsey Theorem for Edges](/theorems/2048). For the inductive step, we mimic the pigeonhole construction from that proof but one level higher: at each stage we pick a new vertex $x_i$ and use the inductive hypothesis on $(r-1)$-subsets to find an infinite set $S_i$ on which the induced colouring $c_{x_i}(F) := c(\{x_i\} \cup F)$ is monochromatic with colour $c_i$. Iterating yields a sequence $(x_i)$ such that the colour of any $r$-subset $\{x_{i_1}, \ldots, x_{i_r}\}$ with $i_1 < \cdots < i_r$ is determined by $c_{i_1}$, so the colour sequence $(c_i)$ takes only two values and pigeonhole on this sequence extracts the required infinite monochromatic subset. [/proofplan]

[step:Reduce to the inductive hypothesis via induction on $r$] We argue by induction on $r \geq 1$. For $r = 1$, any 2-colouring $c : \mathbb{N}^{(1)} \to \{\text{red}, \text{blue}\}$ assigns one of two colours to each singleton $\{n\}$, so by pigeonhole one colour class is infinite; that colour class is the required infinite set $X$ with $X^{(1)}$ monochromatic. For $r = 2$, the statement is precisely the [Infinite Ramsey Theorem for Edges](/theorems/2048). Fix $r \geq 3$ and assume the theorem holds for $r - 1$: for every 2-colouring of $\mathbb{N}^{(r-1)}$ (and of $M^{(r-1)}$ for any infinite $M \subseteq \mathbb{N}$), there is an infinite monochromatic set. [/step]

[step:Build nested infinite sets $S_i$ and vertices $x_i$ using induced $(r-1)$-colourings]We construct, by recursion on $i \geq 1$, a strictly increasing sequence of natural numbers $x_1 < x_2 < \cdots$, a decreasing chain of infinite sets $\mathbb{N} = S_0 \supsetneq S_1 \supsetneq S_2 \supsetneq \cdots$, and a sequence of colours $(c_i)_{i \geq 1} \subseteq \{\text{red}, \text{blue}\}$ with the following property: **(\*) For every $i \geq 1$ and every $F \in S_i^{(r-1)}$, the set $\{x_i\} \cup F$ has colour $c_i$ under $c$.** *Base step* ($i = 1$). Let $x_1 := \min \mathbb{N} = 1$ (or any fixed element of $\mathbb{N}$). Define the induced colouring of $(r-1)$-subsets of $\mathbb{N} \setminus \{x_1\}$ as the map \begin{align*} c_{x_1}: (\mathbb{N} \setminus \{x_1\})^{(r-1)} &\to \{\text{red}, \text{blue}\} \\ F &\mapsto c(\{x_1\} \cup F). \end{align*} This is a well-defined 2-colouring of the $(r-1)$-subsets of the infinite set $\mathbb{N} \setminus \{x_1\}$. By the inductive hypothesis applied to $\mathbb{N} \setminus \{x_1\}$ (which is order-isomorphic to $\mathbb{N}$ and hence admits the same result), there exists an infinite set $S_1 \subseteq \mathbb{N} \setminus \{x_1\}$ with $S_1^{(r-1)}$ monochromatic under $c_{x_1}$. Let $c_1 \in \{\text{red}, \text{blue}\}$ denote this colour. Then (\*) holds for $i = 1$ by definition of $c_{x_1}$. *Inductive step.* Suppose $x_1, \ldots, x_i$, infinite sets $S_1 \supseteq \cdots \supseteq S_i$, and colours $c_1, \ldots, c_i$ have been chosen so that (\*) holds up to index $i$, with $x_j \in S_{j-1}$ for $1 \leq j \leq i$. Pick $x_{i+1} := \min S_i$ (so $x_{i+1} > x_i$ since $x_i \notin S_i$ by the strict-containment property, which we maintain below). Define \begin{align*} c_{x_{i+1}}: (S_i \setminus \{x_{i+1}\})^{(r-1)} &\to \{\text{red}, \text{blue}\} \\ F &\mapsto c(\{x_{i+1}\} \cup F). \end{align*} Since $S_i \setminus \{x_{i+1}\}$ is infinite, the inductive hypothesis yields an infinite set $S_{i+1} \subseteq S_i \setminus \{x_{i+1}\}$ with $S_{i+1}^{(r-1)}$ monochromatic under $c_{x_{i+1}}$; let $c_{i+1}$ be the common colour. Then (\*) holds at index $i+1$, $S_{i+1} \subsetneq S_i$, and $x_{i+1} < \min S_{i+1}$ (so the $x_i$ are strictly increasing).[/step]

[guided]The idea is to iterate the edge-Ramsey strategy one dimension up. In the edge case, we picked a vertex $x_i$ and, using only two possible edge colours, pigeonholed on the colours of the edges from $x_i$ into the remaining infinite set. Here we cannot pigeonhole directly because the colours of $(r-1)$-subsets through $x_i$ are parameterised by an entire family $F$, not by single elements. The fix is to use the inductive hypothesis in dimension $r - 1$ instead of pigeonhole. Concretely, we build by recursion on $i$: - elements $x_1 < x_2 < \cdots$ of $\mathbb{N}$, - a decreasing chain $\mathbb{N} \supsetneq S_1 \supsetneq S_2 \supsetneq \cdots$ of infinite sets with $x_i \in S_{i-1} \setminus S_i$, - colours $c_i \in \{\text{red}, \text{blue}\}$, such that **every $(r-1)$-subset $F$ of $S_i$ satisfies $c(\{x_i\} \cup F) = c_i$**. This is the property (\*) on which the whole argument hinges. *Why this is exactly what we need.* Later, for any indices $i_1 < \cdots < i_r$, we want $c(\{x_{i_1}, \ldots, x_{i_r}\})$ to depend only on $i_1$. Property (\*) applied at index $i_1$ says that $c(\{x_{i_1}\} \cup F) = c_{i_1}$ for every $F \in S_{i_1}^{(r-1)}$. If we can guarantee $\{x_{i_2}, \ldots, x_{i_r}\} \in S_{i_1}^{(r-1)}$, we are done. Since $i_2, \ldots, i_r > i_1$ and the $S_j$ are nested, $x_{i_2}, \ldots, x_{i_r} \in S_{i_1}$ provided each later $x_j$ is chosen inside $S_{j-1}$ (which we arrange by taking $x_{j} \in S_{j-1}$). *Base step ($i = 1$).* Fix $x_1 := 1$. To find $S_1$, we look at $(r-1)$-subsets through $x_1$, i.e. we consider the induced colouring \begin{align*} c_{x_1}: (\mathbb{N} \setminus \{x_1\})^{(r-1)} &\to \{\text{red}, \text{blue}\} \\ F &\mapsto c(\{x_1\} \cup F). \end{align*} This is a 2-colouring of the $(r-1)$-subsets of the infinite set $\mathbb{N} \setminus \{x_1\}$. The inductive hypothesis at level $r - 1$ requires a 2-colouring of $\mathbb{N}^{(r-1)}$, but the statement applies equally to any infinite set (relabelling $\mathbb{N} \setminus \{x_1\}$ order-isomorphically to $\mathbb{N}$ transports the colouring). Applying it, we obtain an infinite $S_1 \subseteq \mathbb{N} \setminus \{x_1\}$ on which $c_{x_1}$ is constant, with value $c_1$. Property (\*) at $i = 1$ is immediate: for $F \in S_1^{(r-1)}$, $c(\{x_1\} \cup F) = c_{x_1}(F) = c_1$. *Inductive step.* Having built $x_1, \ldots, x_i$, $S_1 \supsetneq \cdots \supsetneq S_i$, and $c_1, \ldots, c_i$, we choose $x_{i+1} := \min S_i$. Since $x_i \notin S_i$ (by construction), $x_{i+1} > x_i$. The induced colouring \begin{align*} c_{x_{i+1}}: (S_i \setminus \{x_{i+1}\})^{(r-1)} &\to \{\text{red}, \text{blue}\} \\ F &\mapsto c(\{x_{i+1}\} \cup F) \end{align*} is a 2-colouring of the $(r-1)$-subsets of the infinite set $S_i \setminus \{x_{i+1}\}$. By induction, we obtain an infinite $S_{i+1} \subseteq S_i \setminus \{x_{i+1}\}$ on which $c_{x_{i+1}}$ is constant with value $c_{i+1}$. Note that $S_{i+1} \subsetneq S_i$ (strictly, since $x_{i+1}$ is excluded), and $x_{i+1} \notin S_{i+1}$, so the strict-containment invariant is preserved. Property (\*) at $i+1$ follows.[/guided]

[step:Derive that the colour of $\{x_{i_1}, \ldots, x_{i_r}\}$ depends only on $i_1$] We claim: for all indices $i_1 < i_2 < \cdots < i_r$, \begin{align*} c(\{x_{i_1}, x_{i_2}, \ldots, x_{i_r}\}) = c_{i_1}. \end{align*} Indeed, set $F := \{x_{i_2}, \ldots, x_{i_r}\}$, an $(r-1)$-subset of $\mathbb{N}$. By construction $x_{i_j} \in S_{i_j - 1}$ for each $j$, and the sets $S_k$ are nested decreasing, so $x_{i_2}, \ldots, x_{i_r} \in S_{i_2 - 1} \subseteq S_{i_1}$. Hence $F \in S_{i_1}^{(r-1)}$. Property (\*) at index $i_1$ gives $c(\{x_{i_1}\} \cup F) = c_{i_1}$, which is the claim. [/step]

[step:Pigeonhole on the colour sequence $(c_i)$ to extract the monochromatic infinite set] The sequence $(c_i)_{i \geq 1}$ takes values in the two-element set $\{\text{red}, \text{blue}\}$, so by pigeonhole at least one colour, call it $c^* \in \{\text{red}, \text{blue}\}$, occurs for infinitely many indices. Let \begin{align*} I := \{i \geq 1 : c_i = c^*\}, \qquad X := \{x_i : i \in I\}. \end{align*} Then $I$ is infinite, and since $i \mapsto x_i$ is strictly increasing, $X$ is an infinite subset of $\mathbb{N}$. For any $r$-subset $Y \subseteq X$, write $Y = \{x_{i_1}, \ldots, x_{i_r}\}$ with $i_1 < \cdots < i_r$ and all $i_j \in I$. By the previous step, \begin{align*} c(Y) = c_{i_1} = c^*. \end{align*} Hence $X^{(r)}$ is monochromatic with colour $c^*$, completing the inductive step and the proof. [/step]