Erdoes #70 - working report - Erdős Problem a Day

Erdős problem 70 — wave 5g report

Date: 2026-07-26 UTC

Claim labels used throughout:

[a] elementary-rigorous — proved below from definitions or by a finite certificate whose verification is transparent.
[b] rigorous-modulo-named-theorem — the deduction is rigorous assuming the explicitly cited published theorem.
[c] plausible/structural-unverified — a proposed route or a literature-search conclusion, not a theorem.
[d] computational-only — an exhaustive or SAT computation, not promoted to an infinitary theorem.

0. Mandatory live-page check

I accessed the live page through the Bright Data browser route on 2026-07-26. Direct extraction and a full-page screenshot agreed. The page was last edited 23 January 2026.

Verbatim live statement

> Let $\mathfrak{c}$ be the ordinal of the real numbers, $\beta$ be any countable ordinal, and $2\leq n<\omega$. Is it true that $\mathfrak{c}\to (\beta, n)_2^3$?

Live status, listed result, and collaboration/proof markers

[b] Status badge: OPEN.
[b] Page-listed known result, verbatim: “Erdős and Rado proved that $\mathfrak{c}\to (\omega+n,4)_2^3$ for any $2\leq n<\omega$.”
[b] References displayed by the page: [Er87] and [Va99,7.83].
[b] 0 comments on this problem.
[b] 0 claimed proofs for this problem.
[b] Interested in collaborating: None.
[b] Currently working on this problem: None.
[b] Likes this problem: None.
[b] This problem looks difficult: padillajignacio.
[b] This problem looks tractable: None.
[b] The results on this problem could be formalisable: None.
[b] I am working on formalising the results on this problem: None.
[b] The external-data field says Formalised statement? Yes.

Thus the mandatory stop condition did not fire: there is neither a claimed proof nor a current worker.

The citation popovers identify:

[Er87]: P. Erdős, Some problems on finite and infinite graphs, Logic and Combinatorics (Arcata, 1985), 1987, pp. 223–228, MR 891250.
[Va99]: Some of Paul's favorite problems, booklet for “Paul Erdős and his mathematics”, Budapest, July 1999; the page points to item 7.83.

Live sources: problem page, LaTeX view.

1. Meaning of the relation

[a] On the standard partition-calculus reading,

\[ \mathfrak c\to(\beta,n)^3_2 \]

means that every coloring $f:[\mathfrak c]^3\to\{0,1\}$ has either a set of order type $\beta$ whose triples all have color $0$, or an $n$-element set whose triples all have color $1$. Here $\mathfrak c$ is regarded as its initial ordinal. Cantor's theorem gives

\[ \omega_1\leq \mathfrak c. \]

This simple containment is what permits the published $\omega_1$ theorem below to be transferred to the live problem.

2. Primary-source literature audit

Original problem

[b] The primary Erdős paper exists in the Rényi archive as a six-page scan: Erdős 1987 PDF. On its second printed page (p. 224), Problem 3 records the old $\mathfrak c\to(\omega+n,4)^3_2$ result and asks whether the ordinal and finite targets can be generalized in exactly the direction of the live question.

Stronger published partial result missing from the live remarks

[b] Albin L. Jones proved the following ZFC theorem:

\[ \boxed{\omega_1\to(\omega+\omega+1,n)^3\quad\text{for every }n<\omega.} \tag{J} \]

Primary source: Albin L. Jones, Even more on partitioning triples of countable ordinals, Proc. Amer. Math. Soc. 146 (2018), 3529–3539, DOI 10.1090/proc/13503, official AMS record. I obtained and read the full paper, not just secondary metadata. Its abstract, Proposition 5.2, and final Question 1 all state the displayed relation.

[b] Jones derives (J) in ZFC by first proving the relation under a proper forcing extension satisfying auxiliary axioms and then applying a forcing-absoluteness lemma. I do not reproduce that eleven-page forcing proof here, so every use of (J) is explicitly marked theorem-dependent.

Current-state cross-check

[b] Péter Komjáth's 2025 primary survey, The Erdős–Hajnal problem list, Bull. Symbolic Logic 31 (2025), 418–461, DOI 10.1017/bsl.2025.1, still calls

\[ \omega_1\to(\alpha,n)^3\qquad(\alpha<\omega_1,\ n<\omega) \]

open and records Jones's $\omega+\omega+1$ theorem. This agrees with Jones's own final remarks and with the live page's 2026 OPEN badge.

[c] Exact-formula searches, DOI/citation searches, and inspection of the 2025 survey found no later primary source improving (J). This is an honest search result, not a proof that no uncatalogued result exists. The live page itself warns that its remarks may omit relevant literature; in this case they do omit Jones's published strengthening.

3. A rigorously settled region of the live problem

The partial theorem

For every countable ordinal $\beta$ and every $2\leq n<\omega$:

1. [a] If $n\leq 3$, then $\mathfrak c\to(\beta,n)^3_2$.

2. [b] If $\beta\leq\omega+\omega+1=\omega\cdot2+1$, then

$\mathfrak c\to(\beta,n)^3_2$ for every finite $n$.

Consequently, the portion not covered by these two statements is exactly

\[ \boxed{4\leq n<\omega,\qquad \omega\cdot2+2\leq\beta<\omega_1.} \tag{R} \]

The word “exactly” here means “exactly the parameter region left after the two proved partial statements”; it does not assert that every pair in (R) is independently known open in the literature.

Proof of the elementary $n=2,3$ cases

[a] For $n=2$, every two-element set is color-1 homogeneous because it has no three-element subsets.

[a] For $n=3$, let $f:[\mathfrak c]^3\to2$. If some triple has color $1$, its three vertices are the required color-1 set. If no triple has color $1$, every triple has color $0$. Since $\beta<\omega_1\leq\mathfrak c$, the initial segment $\beta\subseteq\mathfrak c$ is then a color-0 set of order type $\beta$.

Transfer of Jones's theorem to $\mathfrak c$

[b] Fix $f:[\mathfrak c]^3\to2$, and restrict it to $[\omega_1]^3$. By (J), either:

there is $H\subseteq\omega_1$ of order type $\omega+\omega+1$ all of whose triples have color $0$; or
there is an $n$-element color-1 set.

In the first case, for every $\beta\leq\omega+\omega+1$, the appropriate initial segment of $H$ has order type $\beta$ and remains color-0 homogeneous. This proves the second partial statement.

This is a genuine strengthening of the range displayed on the live page: the finite target is arbitrary, and the ordinal target reaches $\omega\cdot2+1$.

4. The first open cell and the exact proof-route obstruction

[b] Jones explicitly identifies

\[ \omega_1\to(\omega+\omega+2,4)^3 \tag{F} \]

as the simplest open instance of the $\omega_1$ conjecture. Relation (F) would imply the corresponding $\mathfrak c$ relation by restriction, just as above.

What the 2018 proof supplies

In the auxiliary model used in Jones's proof, Lemma 4.2 produces, when the color-1 finite alternative is absent, sets

\[ A\in[\omega_1]^\omega,\qquad B\in[\omega_1]^{\omega_1},\qquad Awith

\[ [A,B]^{2,1}\subseteq f^{-1}(0). \tag{one side} \]

Here $[A,B]^{p,q}$ denotes triples having $p$ points in $A$ and $q$ points in $B$. The later “good pair” argument recovers enough additional compatibility for one final apex $\beta$, producing two $\omega$-blocks followed by one point—exactly $\omega+\omega+1$.

A precise sufficient missing lemma

Consider the following strengthening for fixed $n\geq4$:

\[ \begin{split} T_n:\quad&\text{If }f:[\omega_1]^3\to2\text{ has no color-1 }n\text{-set, then}\\ &\text{there exist }A\in[\omega_1]^\omega,\ B\in[\omega_1]^{\omega_1}, \ A[c] I do not prove $T_n$, and do not claim it is true. It cleanly names the extra cross-orientation that the present machinery lacks.

[a] If $T_n$ holds, then

\[ \omega_1\to(\omega\cdot2+2,n)^3. \]

Proof: infinite Ramsey on $[A]^3$ yields an infinite homogeneous

$H\subseteq A$. It cannot have color $1$, since it would contain a

color-1 $n$-set. Choose the first $\omega$ elements of $H$, in their

inherited well-order, and call this color-0 set $A'$; then

$\operatorname{otp}(A')=\omega$. The uncountable

$B\subseteq\omega_1$ has order type $\omega_1$. Apply Jones's theorem

inside $B$ with ordinal target $\omega+2$. Again the color-1 alternative

is excluded, so obtain a color-0 $C\subseteq B$ of order type

$\omega+2$. Both cross orientations are color $0$ by $T_n$, hence

\[ [A'\cup C]^3\subseteq f^{-1}(0),\qquad \operatorname{otp}(A'\cup C)=\omega+(\omega+2)=\omega\cdot2+2. \]

Thus, for this proof route, the missing mathematical content is not another finite Ramsey bound: it is simultaneous control of $[A,B]^{1,2}$ in addition to the already obtained $[A,B]^{2,1}$. Jones's one-apex maximal-good-pair argument does not provide that second rectangle.

5. Reproducible finite computation

The standalone checker is:

runs/erdos70_wave5g_verify.py

It uses only the Python standard library.

Explicit finite certificate

[a] The script contains a fixed 220-bit coloring of $[12]^3$, in lexicographic triple order, and directly checks all

\[ \binom{12}{4}=495 \]

four-sets. Every four-set contains triples of both colors. Therefore the finite three-uniform Ramsey number obeys

\[ R_3(4,4)\geq13. \]

The certificate has 110 triples of each color. Its ordered bit string has SHA-256

c2ef875585d473842d8d2835da204575246093104b3d2ee7e698140bf03ee033.

The assignment was discovered by SAT, but certificate verification is a from-scratch exhaustive loop independent of the SAT package.

Exact labeled census

[d] Exhaustive enumeration gives the following exact counts of two-colorings of $[N]^3$ with no monochromatic four-set:

|---:|---:|---:|---:|---:|

| 3 | 1 | 2 | 2 | $1$ |

| 4 | 4 | 14 | 16 | $7/8$ |

| 5 | 10 | 512 | 1024 | $1/2$ |

| 6 | 20 | 118784 | 1048576 | $29/256$ |

The script also recomputes and checks the complete histogram by number of color-1 triples; those data are not needed for the ordinal deduction.

Independent check of Jones's finite gadget

Jones's Claim 4.2.3 uses

\[ x_{n,k}=[0,n-k)\cup[kn,kn+k)\qquad(0\leq k\leq n). \]

[a] For \(i \[ \begin{aligned} u&=[0,n-j),\\ v_0&=[n-j,n-i)\cup[in,in+i),\\ v_1&=[jn,jn+j). \end{aligned} \]

Then $x_{n,i}=u\cup v_0$, $x_{n,j}=u\cup v_1$,

$|v_0|=|v_1|=j$, and \(u

$x_{n,i}\mathrel{\triangleleft}x_{n,j}$ in Jones's notation. Moreover, for

\(i \[ |x_{n,i}\cap x_{n,j}|=n-j>n-k=|x_{n,j}\cap x_{n,k}|. \]

This algebraically verifies the finite gadget for every $n$.

[d] As a defense against transcription or boundary mistakes, the script independently checks every indexed pair and triple for $1\leq n\leq50$: 22,100 pair checks and 270,725 triple checks.

Reproduction log

Command:

python runs/erdos70_wave5g_verify.py

Observed on Python 3.12.3:

certificate N=12: PASS; checked 495 four-sets; 110 triples of each colour; hence R_3(4,4) >= 13
certificate bit-string sha256: c2ef875585d473842d8d2835da204575246093104b3d2ee7e698140bf03ee033
exact finite census (labelled vertices, two named colours):
N  C(N,3)  avoiding/total  fraction
3       1         2/2         1
4       4        14/16        7/8
5      10       512/1024      1/2
6      20    118784/1048576   29/256
Jones x_(n,k) gadget: PASS for 1 <= n <= 50; checked 22100 ordered-index pairs and 270725 index triples
ALL CHECKS PASS

Measured runtime was 1.96 seconds, peak RSS 16,172 KiB. Checker SHA-256:

a8380f31b6d7031b78c8a70a42c95927466a296d2dc644a1c35733dc2c23aee0.

6. Why computation and standard machinery stop here

[a] A finite ordered set cannot distinguish the target order types

$\omega$, $\omega+1$, $\omega\cdot2+1$, and

$\omega\cdot2+2$. Consequently, no table of finite hypergraph Ramsey

numbers supplies the missing successor point in (F).

[a] Finite Ramsey guarantees arbitrarily large finite homogeneous sets, but

those sets need not form a nested branch assembling into a homogeneous set of a

specified countable order type. This is the failed compactness/uniformity step.

[d] The raw finite search space at $N$ is

$2^{\binom N3}$: already $2^{220}$ at $N=12$ and $2^{286}$ at

$N=13$. SAT readily produced the 12-vertex certificate. An exploratory

unsatisfiability run at $N=13$ did not finish within 120 seconds and was

terminated, so no finite upper-bound claim is made here.

[a] More CPU time on those finite instances would not decide (F), regardless

of whether the exact finite value were certified. The required advance is a

transfinite structural lemma—one possible sufficient form is the two-sided

rectangle $T_n$—or an explicit uncountable coloring giving a counterexample.

There is therefore no honest finite core-hour estimate for “searching farther”

that would close the live problem.

7. Bottom line

[a] The cases $n=2,3$ hold for every countable $\beta$.
[b] Jones's verified 2018 theorem implies the live relation for every finite $n$ and every $\beta\leq\omega\cdot2+1$, a substantial published range missing from the live remarks.
[a] The remaining parameter region is (R).
[b] The first unresolved $\omega_1$ instance is $(\beta,n)=(\omega\cdot2+2,4)$.
[c] The exact missing feature in Jones's proof route is two-sided, rather than one-sided, cross-triple homogenization.
[d] The supplied checker gives an explicit finite certificate, an exact small census, and an independent audit of the key finite gadget, but none is misrepresented as resolving the transfinite uniformity step.

PARTIAL: live page is OPEN with no worker/claim; verified $n\leq3$ for all countable $\beta$ and, modulo Jones 2018, every finite $n$ for $\beta\leq\omega\cdot2+1$, leaving $n\geq4,\ \beta\geq\omega\cdot2+2$, with a reproducible finite certificate and a precise sufficient two-sided-rectangle lemma isolating this proof route's missing cross orientation.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger