Bohr protected preparations

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One counterexample attempt sought finite sieves with all moduli at most $N$, very small full survivor measure, but a positive proportion of survivors across $[N,N^K]$, for fixed $K>1$. Such a preparation alone would not yet be a counterexample: it would still need to be incorporated into compatible infinite singleton blocks producing fixed logarithmic overdeletion. No such preparation or infinite construction was obtained.

The following theorem rules out protecting a single interval under a rotation as the proposed preparation, even when the slope changes arbitrarily with $N$. It is a finite counting estimate, not a general density theorem.

Theorem

Fix an interval $I\subset\mathbb R/\mathbb Z$ of length $\ell>0$. Let

$$S=\{1\le n\le H:\{n\theta\}\in I\}.$$

Suppose a finite family has distinct moduli $q\le N$, one residue class per modulus, and every full class avoids $S$. Let $\delta$ be the full periodic survivor density. Then

$$\delta\ge\frac{|S|}{H} -C_\ell\sqrt{\frac{N\log(2N)}H} \qquad(H\ge C_\ell N\log(2N)). \tag{1}$$

The constant is independent of $\theta$, the residues, and the family.

The same assertion holds after translating the interval of integers and the Bohr interval. Thus on $[N,N^K]$, $K>1$ fixed, a protected Bohr set of positive counting proportion forces a positive full survivor measure. This includes rational, Liouville, and $N$-dependent slopes. It does not treat arbitrary protected sets.

An avoided progression yields a rational approximation

Put $J=\lceil8/\ell\rceil$. If a progression modulo $q\le N$ avoids $S$, its rotation phases form at least $H/q-1$ consecutive terms with step $q\theta$, all avoiding $I$. The elementary pigeonhole approximation lemma gives a $1\le j\le J$ with

$$\|jq\theta\|\le1/(J+1)<\ell/4.$$

For clarity, the lemma follows by putting $0,q\theta,\ldots,Jq\theta$ on the circle and choosing two of the $J+1$ points at circular distance at most $1/(J+1)$.

The corresponding $j$-spaced subsequence has length at least $H/(jq)-2$. If its signed step $\xi$, with $|\xi|=\|jq\theta\|<\ell/4$, travels a full circle, consecutive points leave gaps smaller than $\ell$, contradicting avoidance. For $H\ge8JN$, this gives

$$\|jq\theta\|\le A jq/H$$

with an absolute constant $A$, for example $A=8$. The assertion also holds if the step is zero. After reduction, there is consequently a rational $a/b$ satisfying

$$b\mid jq,\qquad b\le JN,\qquad |\theta-a/b|\le A/H. \tag{2}$$

We may replace $\theta$ by its fractional part throughout.

Let $\mathcal A$ be all reduced fractions satisfying the last two conditions in (2). If no forbidden class is present, (1) is immediate. Otherwise $\mathcal A$ is nonempty. Define

$$\eta=\sqrt{\frac{N\log(2N)}H}, \qquad B=\frac{\log(2N)}\eta =\sqrt{\frac{H\log(2N)}N},$$

and let $b_0$ be the least denominator in $\mathcal A$. It suffices to work when $\eta$ is smaller than a sufficiently small constant depending on $\ell$: the other range follows from the trivial bounds $\delta\ge0$ and $|S|/H\le1$ after increasing $C_\ell$.

No small denominator: the whole sieve has small cost

Suppose $b_0>B$. Fractions with denominators in $[Q,2Q)$ are separated by at least $1/(4Q^2)$. Since the approximation interval has length $2A/H$, their number in $\mathcal A$ is $O(1+Q^2/H)$.

For a fixed reduced denominator $b$, condition $b\mid jq$ for some $j\le J$ implies

$$\sum_{\substack{q\le N\\b\mid jq\text{ for some }j\le J}}\frac1q \le\sum_{j\le J}\frac{(b,j)}b H_{\lfloor N(b,j)/b\rfloor} \ll_\ell\frac{\log(2N)}b.$$

Summing this bound over the approximants, in dyadic denominator blocks from $b_0$ to $JN$, gives

$$\sum_{q\text{ selected}}\frac1q \ll_\ell\log(2N)\left(\frac1{b_0}+\frac NH\right) \ll_\ell\eta+\eta^2\ll_\ell\eta. \tag{3}$$

Distinctness and one class per modulus now matter: the full Haar union bound gives $\delta\ge1-O_\ell(\eta)$, which implies (1).

A small denominator: a common periodic region survives

Suppose $b_0\le B$, and choose its reduced approximant $a/b$. It is the only fraction in $\mathcal A$ when $\eta$ is sufficiently small. Indeed any distinct one with denominator $b'\le JN$ has distance at least $1/(bJN)$ from it, whereas two approximants are at distance at most $2A/H$. The inequality $2ABJN/H=2AJ\eta<1$ rules this out.

Write

$$\theta=a/b+\varepsilon,\qquad |\varepsilon|H\le A, \qquad L=bN.$$

We have $L/H\le BN/H=\eta$, so $L<H/4$ in the relevant range.

If $\varepsilon=0$, the set $\{n:\{an/b\}\in I\}$ is $b$-periodic. Any forbidden class intersecting it would have a common point with it in $[1,H]$, because the combined period is at most $bN=L<H$. Thus the whole periodic set survives. Its density differs from $|S|/H$ by at most $b/H$, giving (1).

Assume now $\varepsilon\ne0$, and let $w=|\varepsilon|L$. For $L\le t\le H-L$, define

$$R_t=\left\{r\bmod b: \{ar/b+\varepsilon t\}\text{ lies in }I \text{ at distance greater than }w\text{ from its boundary}\right\}.$$

Every forbidden full class avoids this entire $b$-periodic region. Otherwise CRT supplies a common integer $n$ within distance $L$ of $t$, and inside $[1,H]$, since the combined period is at most $bq\le L$. Its actual phase differs from $ar/b+\varepsilon t$ by at most $|\varepsilon|L=w$, so $n\in S$, a contradiction. Consequently

$$\delta\ge\max_{L\le t\le H-L}|R_t|/b. \tag{4}$$

Here is the averaging estimate required for (4). For each residue $r\bmod b$, the function

$$f_r(t)=\mathbf1_I(\{ar/b+\varepsilon t\})$$

is constant on at most $O_A(1)$ intervals in $[0,H]$, since the total phase travel is at most $A$. Counting on the progression $n\equiv r\pmod b$ therefore differs from $b^{-1}\int_0^H f_r(t)\,dt$ by $O_A(1)$. Summing over $r$ gives

$$|S|=\int_0^H\frac1b\sum_r f_r(t)\,dt+O_A(b).$$

Eroding $I$ by the margin $w$ removes, for each $r$, at most $O_A(w/|\varepsilon|)=O_A(L)$ units of time: the phase passes each endpoint only $O_A(1)$ times. Deleting the two boundary time intervals also costs at most $2L$. Hence

$$\int_L^{H-L}\frac{|R_t|}{b}\,dt\ge |S|-O_A(L).$$

This reasoning permits either choice of endpoint inclusion in $I$; for nonzero $\varepsilon$, possible endpoint visits add only $O_A(1)$ counting errors per residue. Combining the last display with (4) yields

$$\delta\ge |S|/H-O_A(L/H) \ge |S|/H-O_\ell(\eta),$$

which proves (1).

Interpretation for the attempted construction

On a window entirely above all preparatory moduli, delayed avoidance is full congruence avoidance. Translating $[N,N^K]$ to an initial interval changes neither moduli nor full periodic density; it only translates the Bohr interval. Applying (1) with $H=\lfloor N^K\rfloor-N$ makes its error tend to zero. Thus a positive-density protected single Bohr interval cannot coexist with a preparatory survivor measure tending to zero.

The theorem neither controls high-complexity protected sets nor proves that every finite sieve with small Haar survivor measure has few survivors on such a window. The general preparation and the infinite counterexample remain unconstructed.