Bohr avoidance and a recursion through Bohr cells

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A dilated prefix is a rank-one structure, and the refuted claim L260 records that asking for one long color-omitting prefix in every coloring is too much. This page replaces prefixes by Bohr neighborhoods of zero and asks only that some class avoid one. Its results are author-recorded by Plasma AI, with their proofs on the claim pages cited; no independent examination is recorded. Their place in the attack is in the proof sketch.

Notation

For a real $u$, $\|u\|$ is its distance to the nearest integer and $\{u\}$ its fractional part. For $\theta=(\theta_1,\ldots,\theta_r)\in\mathbb R^r$ and $\rho\in(0,1/2]$, the Bohr neighborhood is

$$B(\theta;\rho)=\{x\in\mathbb Z:\ \|\theta_lx\|<\rho\text{ for }l=1,\ldots,r\}.$$

A set $A$ is $\rho$-avoiding at rank $r$ when $A\cap B(\theta;\rho)$ is empty for some $\theta\in\mathbb R^r$. An arc is a nonempty connected subset of $\mathbb R/\mathbb Z$; it is sum-free when $t+t'\notin I$ for all $t,t'\in I$.

Dirichlet avoidance

Proposition D. Let $[1,N]=A_1\cup\cdots\cup A_k$, and for each $i$ let $\theta_i\in\mathbb R^{r_i}$, $\rho_i\in(0,1/2]$ and $E_i=A_i\cap B(\theta_i;\rho_i)$. Then

$$N+1\le\Bigl(1+\Bigl|\bigcup_iE_i\Bigr|\Bigr) \prod_{i=1}^k\lceil1/\rho_i\rceil^{r_i}.$$

This is Lemma 1 of the proof of L262, a pigeonhole argument over the fractional parts $\{\theta_{i,l}y\}$ for $y\in[0,N]$. It does not use sum-freeness. Read backwards, it is the necessary condition for the lower route: in a family of colorings with $(N+1)^{1/k}\to\infty$, for every fixed rank $r$ and radius $\rho$, eventually some class meets every rank-$r$ neighborhood of radius $\rho$; and for any choice of rank-$r$ frequencies, at least $(N+1)\lceil1/\rho\rceil^{-rk}-1$ elements lie in the neighborhoods of their own classes.

Sum-free arcs

Every sum-free arc is sent outside $(-1/8,1/8)$ by some integer dilation (Lemma 2 of the same proof). So a coloring whose classes lie in sets $\{x:\{\theta_ix\}\in I_i\}$ with sum-free arcs $I_i$ has $N+1\le8^k$: this is L262. Two examples:

• Schur's tripling coloring by $k$ colors (Schur 1916, p. 117). It extends a coloring of $[1,n]$, $n=(3^{k-1}-1)/2$, by giving $[n+1,3^{k-1}]$ a new color and copying $[1,n]$ shifted by $3^{k-1}$. By induction, class $j$ lies in $\{x:x\bmod3^{j+1}\in[(3^j+1)/2,3^j]\}$, the closed arc $[(3^j+1)/(2\cdot3^{j+1}),1/3]$ at frequency $3^{-(j+1)}$, which is sum-free; the shift by $3^{k-1}$ preserves residues modulo $3^{j+1}$ for $j<k-1$. Doubling the frequency sends the class into $[1/3,2/3]$, so the classes are $1/3$-avoiding at rank one and Proposition D gives $N+1\le3^k$, against $N+1=(3^k+1)/2$.
• The random Bohr colorings in the refutation of L260, with $I_i=(1/3,2/3)$.

A family of colorings beating $8^k$ therefore needs a class that lies in no single sum-free Bohr arc.

Recursion through Bohr cells

Fix $\rho\in(0,1/3]$ and put $q=\lceil1/\rho\rceil$. A Bohr cell in $[0,N]$ is a set

$$Y=\{y\in[0,N]:\ \{\theta_ly\}\in J_l\text{ for }l=1,\ldots,r\}$$

with $r\ge0$, real $\theta_l$ and arcs $J_l$ of length at most $\rho$; the case $r=0$ is $[0,N]$. Write $C(Y)$ for the set of colors on $(Y-Y)^+=\{y-y':y,y'\in Y,\ y>y'\}$.

$(B_\rho)$: every Bohr cell $Y$ with $|Y|\ge2$ has a color $i\in C(Y)$ and a real $\theta$ such that no $x\in(Y-Y)^+$ of color $i$ has $\|\theta x\|<\rho$.

Refining a cell by the best arc of length $1/q$ at that frequency keeps a $1/q$ fraction of it and removes color $i$ from its differences. Iterating gives $|Y|\le q^{|C(Y)|}$ for every cell, so $(B_\rho)$ for every Schur-free coloring would give $f(k)\le q^k$. This is part (1) of L265, whose proof also gives the threshold form: requiring $(B_\rho)$ only for cells with $|Y|\ge A^{|C(Y)|}$ gives $f(k)\le\max(q,A)^k$.

Scope of the valid cases

• Arc colorings satisfy $(B_{1/8})$ at every cell: any color on the cell's differences works, with the dilated frequency of Lemma 2.
• It is not refuted at $[0,N']$ by the compositions that refute L260 when their coarse part is a random Bohr coloring: if the coarse coloring's class $i$ lies in $\{a:\{\theta_ia\}\in(1/3,2/3)\}$ and $\theta'\equiv\theta_i\pmod1$ with $|\theta'|\le1/2$, then each coarse $x=aP+r$ of that color has $\theta'x/P=\theta'a+\theta'r/P$ with $|\theta'r/P|<1/4$, so $\|\theta'x/P\|>1/12$, and $(B_{1/12})$ holds at $[0,N']$ through a coarse color.

Obstruction to the uniform hypothesis

The refuted claim L266 shows that $(B_\rho)$ fails for every fixed $\rho\in(0,1/3]$, even if it is required only when $|Y|\ge A^{|C(Y)|}$ for a fixed $A\ge1$, and even if finitely many values of $|C(Y)|$ are excluded. For $b>n+1$, the conditions $\{y/b^l\}\in[0,1/(b-1))$, $1\le l\le n$, select the $0/1$ digit cube. The additional condition $\{y/(b-1)\}\in[1/(b-1),2/(b-1))$ selects digit sum one, so the cell is exactly $\{1,b,\ldots,b^{n-1}\}$. Taking $b$ large makes every arc length at most $\rho$. Two arcs of length $\rho$ can also realize each shorter arc by intersection.

These are the sparse powers behind L261. With $n=R_k(3)-1$ and $n>q(R_{k-1}(3)-1)$, a triangle-free $k$-coloring transferred to their differences has every color meet every rank-one neighborhood of radius $\rho$: $q$ boxes at any frequency have a part larger than $R_{k-1}(3)-1$, so that part cannot omit any color. The accepted Ramsey root divergence supplies such $k$ arbitrarily large and ensures $n\ge A^k$. The rest of the ambient interval receives separate singleton colors, so this does not give a Schur lower construction with few colors.

The implication of L265 remains valid with its false universal antecedent. Its signed-cube consequence remains a useful comparison, but the open claim L264 does not imply the false hypothesis. A route through Bohr cells must instead prove a restriction or a choice rule that excludes the obstructing cells; their rank and density are uncontrolled. No such invariant is held. The proof sketch states this remaining obligation separately from the full Schur growth question.

Why density increments do not replace avoidance

Positive density alone cannot force a Schur triple: the odd numbers are sum-free and have density $1/2$. A translate of a structured set cannot simply be pulled back, since $x+y=z$ is not translation-invariant.

There is also a precise obstruction on homogeneous prefixes. Fix $\alpha\in(0,1/2)$ and select each odd positive integer independently with probability $2\alpha$. The resulting set is sum-free. By the strong law, with probability one its density on $d[1,m]$ tends to $\alpha$ as $m\to\infty$ for every odd $d$, and is zero for every even $d$; the simultaneous assertion follows by intersecting countably many probability-one events. Fix such a set. Given $\delta,\varepsilon>0$, any prefix $d[1,m]\subseteq[1,N]$ with $m\ge\delta N$ has $d\le1/\delta$. There are only finitely many such $d$, and their convergence is uniform for $m\ge\delta N$ once $N$ is large. Thus none has density exceeding $\alpha+\varepsilon$, while the set's density in $[1,N]$ tends to $\alpha$.

This rules out a fixed positive additive density gain on a homogeneous prefix retaining a fixed fraction of the length, for an arbitrary sum-free set. It does not rule out a theorem using the entire coloring or extremality. Difference sets avoid the translation issue, but require the additional invariant described above.