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For an infinite cardinal $\rho$ put ${\beth }_0(\rho )=\rho$ and ${\beth }_{n+1}(\rho )=2^{{\beth }_n(\rho )}$. The Erdős-Rado theorem for finite arities isThe partition relation requires a homogeneous subset of order type ${\rho }^{+}$ for every coloring of $(n+1)$-element subsets by at most $\rho$ colors. In particular it gives uncountable homogeneous sets. The case $n=0$ is the infinite pigeonhole principle: a union of $\rho$ sets each of size at most $\rho$ cannot have size ${\rho }^{+}$.

We give the induction step by a closed elementary-submodel construction of an end-homogeneous sequence. Let $\mu$ be infinite, $\theta =(2^{\mu }{)}^{+}$, and $c:[\theta {]}^{r+1}\to \rho$ with $\rho \le \mu$. Choose a sufficiently large regular $\chi$ and an elementary submodel $M\prec H_{\chi }$ with $c,\theta \in M$, all ordinals below ${\mu }^{+}$ in $M$, $|M|=2^{\mu }$, and closure under externally given sequences of length at most $\mu$. To construct it, take ${\mu }^{+}$ successive elementary Skolem-hull closures, at each step including all such sequences from the preceding stage. Cardinality stays $2^{\mu }$ because $(2^{\mu }{)}^{\mu }=2^{\mu }$. In the final union every sequence of length at most $\mu$ has all its entries in some earlier stage, by regularity of ${\mu }^{+}$, and so was included at the next stage.

Put $\beta =\sup (M\cap \theta )$. Regularity of $\theta$ gives $\beta <\theta$, and $\beta \notin M$ since otherwise its successor would contradict the definition of the supremum. Recursively, for $\alpha <{\mu }^{+}$ choose $x_{\alpha }\in M\cap \theta$ above all earlier $x_{\xi }$ and satisfyingThere are at most $\mu$ constraints. Their parameters and their colors belong to $M$, and closure puts their complete code in $M$. The point $\beta$ witnesses their simultaneous satisfiability in $H_{\chi }$; elementarity supplies a witness in $M$. This gives an increasing end-homogeneous sequence of length ${\mu }^{+}$: on it, the color of an $(r+1)$-tuple depends only on the first $r$ entries.

For the induction from arity $n$ to $n+1$, take $\mu ={\beth }_{n-1}(\rho )$ and $r=n$. Color the $n$-tuples of the resulting sequence by their common color with a later point, equivalently by their color with $\beta$. Its index order has type ${\mu }^{+}={\beth }_{n-1}(\rho {)}^{+}$, so the induction hypothesis supplies a homogeneous index set of type ${\rho }^{+}$. The original $(n+1)$-tuples on those indices have that same color. This completes the proof without substituting the theorem's name for its induction.

A useful ordinal partition-bound function isIt is total: choose an infinite cardinal $\rho \ge |\alpha |$ with ${\rho }^{+}>\alpha$; the pairs case supplies $(2^{\rho }{)}^{+}$ with the required homogeneous order type for all these color numbers. It is nondecreasing and satisfies $f(\alpha )\ge \alpha$. Start iterating $f$ and take a supremum $\kappa$ at a limit stage. If the increasing iterates do not attain that supremum, then for every $\alpha <\kappa$ some iterate $a_{\xi }$ satisfies $\alpha \le a_{\xi }<\kappa$, andIf the iterates have already stabilized, their supremum is a fixed point directly. Thus the relevant nontrivial case is an uncountable cardinal $\kappa$ closed below itself under $f$.

Such closure implies that $\kappa$ is a strong limit. Suppose $\lambda <\kappa$ is infinite and $\kappa \le 2^{\lambda }$. Take distinct binary strings of length $\lambda$, indexed in order by $\kappa$, and color a pair by its first differing coordinate. No three strings are homogeneous: three bits cannot be pairwise different at the same coordinate. Closure would give $f(\lambda +1)<\kappa$, and its restriction would have a homogeneous subset of order type $\lambda +1$ in $\lambda$ colors, a contradiction. Hence $2^{\lambda }<\kappa$; finite exponents are harmless for an uncountable $\kappa$.

Use the usual tree property: every set-theoretic tree of height $\kappa$ with nonempty levels of cardinality below $\kappa$ has a cofinal branch. For an uncountable cardinal satisfying this all-set-theoretic trees convention, $\kappa$ must be regular. If it were singular with cofinality $\tau <\kappa$, attach to one root $\tau$ chains whose lengths increase cofinally to $\kappa$. Every level has at most $\tau$ nodes, the height is $\kappa$, and no branch is cofinal. This contradiction proves regularity. Combined with the preceding closure argument it proves that the cardinal under discussion is strongly inaccessible.

For any $c:[\kappa {]}^2\to \nu$, $0<\nu <\kappa$, make a separated set-theoretic tree of functions. Its level $\alpha$ consists of the functions $t:\alpha \to \nu$ realized as $t(\xi )=c(\{\xi ,\beta \})$ for some $\beta \ge \alpha$, with $\beta <\kappa$. Order these functions by restriction. Levels are nonempty and have size at most ${\nu }^{|\alpha |}<\kappa$, using the strong-limit property. The tree property gives a branch $b:\kappa \to \nu$. Recursively select $x_{\eta }$ for $\eta <\kappa$, larger than all earlier selections, whose coloring function agrees with $b$ through a level above those earlier selections; regularity allows that level to remain below $\kappa$, and its node has a realizing point. Then $c(\{x_{\xi },x_{\eta }\})=b(x_{\xi })$ for $\xi <\eta$. One of the fewer-than-$\kappa$ colors occurs on $\kappa$ selected points, by regularity. These points are homogeneous, proving $\kappa \to (\kappa {)}_{\nu }^2$ for every $\nu <\kappa$. ConsequentlyA strictly increasing countable iteration has a supremum of countable cofinality, so it cannot meet this regular set-theoretic tree-property condition; no continuity of $f$ has been assumed.

The final printed assertion needs its convention specified. The usual tree property alone does not imply strong inaccessibility; it can consistently hold at successor cardinals. In the argument just given, strong inaccessibility follows from function closure together with that set-theoretic tree property. There is also a stronger branching form of the tree property for which the assertion holds literally: require a cofinal branch in every separated rooted height-$\kappa$ set-theoretic tree whose nodes each have fewer than $\kappa$ immediate successors, without restricting level sizes. The singular-chain example proves regularity under this convention too. If $2^{\lambda }\ge \kappa$ for $\lambda <\kappa$, take the full binary set-theoretic tree through level $\lambda$, choose $\kappa$ distinct nodes on that level, and attach to its $\xi$th chosen node a chain of length $\xi$, for $\xi <\kappa$. It has height $\kappa$ and at most two successors per node, but every branch has length below $\kappa$. Thus this branching property implies the strong-limit condition and hence strong inaccessibility. At a strongly inaccessible cardinal it agrees, for separated set-theoretic trees, with the usual level-size convention: regularity bounds the union of fewer than $\kappa$ successor sets, and strong-limit arithmetic bounds the possible predecessor sequences at limit levels.