Prime block obstructions

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The constructions here defeat proposed intermediate lemmas. Neither gives private positive integer endpoints or a counterexample to Problem 25.

Positive density does not yield a scaled Behrend subfamily

Call $B$ Behrend when its full multiples union has Haar measure one; by the established multiples theorem this is equivalent to logarithmic density one for that union.

Theorem. For every $0<\eta<1$ there is a set $D$ of natural density greater than $1-\eta$ such that, for every integer $d\ge1$, the set

$$B_d=\{b\ge2:db\in D\}$$

is not Behrend. In particular no subfamily of $D$ of the form $dB$, with $1\notin B$, can have $B$ Behrend.

Proof. Choose disjoint finite prime blocks $P_j$ such that

$$\varepsilon_j=\prod_{p\in P_j}(1-1/p)>0, \qquad\sum_j\varepsilon_j<\eta.$$

Successive finite blocks exist by divergence of the prime reciprocal sum. Put $L_j=\prod_{p\in P_j}p$ and choose increasing integers $N_j\ge L_j/\varepsilon_j$. Define

$$D=\{n\ge N_1:(n,L_j)>1\text{ whenever }N_j\le n\}.$$

Let $S_k$ be the periodic set satisfying the first $k$ block conditions. CRT gives $d(S_k)=\delta_k=\prod_{j\le k}(1-\varepsilon_j)$, and $D$ is eventually contained in $S_k$. If $X\ge N_j$, the number of integers at most $X$ failing block $j$ is at most

$$\varepsilon_jX+L_j\le2\varepsilon_jX.$$

Thus

$$\underline d(D)\ge\delta_k-2\sum_{j>k}\varepsilon_j, \qquad\overline d(D)\le\delta_k.$$

Letting $k\to\infty$ proves $d(D)=\delta=\prod_j(1-\varepsilon_j)>1-\eta$.

Fix $d$ and choose a block $j$ with $(d,L_j)=1$. The exceptional set

$$F=\{b\in B_d:(b,L_j)=1\}$$

is finite, because every member has $db<N_j$. It excludes $1$, so $\mu(M(F))<1$: integers coprime to the finitely many prime factors of $F$ avoid $M(F)$. Every other member of $B_d$ has a prime factor in $P_j$. Consequently

$$M(B_d)\subseteq M(F)\cup M(P_j).$$

The two sets on the right depend on disjoint prime coordinates, since each member of $F$ is coprime to $L_j$. Their union has measure

$$1-(1-\mu(M(F)))\varepsilon_j<1.$$

This proves the theorem. $\square$

A positive-measure tail need not contain a divisibility cylinder

The thresholds may also be chosen so that $N_{j+1}>\prod_{i\le j}\max P_i$. Then

$$\bigcap_{Y\ge1}\bigcup_{q\in D,\ q>Y}q\widehat{\mathbb Z} =F_\infty:=\{z:\text{each }P_j\text{ contains a prime dividing }z\}.$$

One inclusion follows from the eventual block requirements on $D$. For the other, choose a dividing prime $p_i\in P_i$ for each $i$ and put $d_j=\prod_{i\le j}p_i$. These grow without bound, satisfy $d_j<N_{j+1}$, and eventually belong to $D$, since every condition active at $d_j$ has index at most $j$. They all divide $z$. The event $F_\infty$ has measure $\delta>0$, but for any $d$ a fresh block coprime to $d$ has positive conditional probability of being missed on $d\widehat{\mathbb Z}$. Thus no divisibility cylinder is contained in $F_\infty$, even up to a null set.

Marginal bounds do not control conditional survivor fractions

A proposed endpoint estimate used a finite period $P$, a survivor set $S\subseteq\mathbb Z/P\mathbb Z$, and

$$f(q,n)=\frac{|S\cap[n]_{(q,P)}|}{P/(q,P)}.$$

Even if a measure $\gamma$ on pairs $(q,n)$ has both marginals bounded by uniform measure, has fixed mass $1/2$, and is supported on $n\in S$ with $|S|/P=1/2$, the integral $\int f\,d\gamma$ can tend to zero.

Take $P$ to be the product of the primes up to a growing bound. Write

$$K(n)=|\{p\mid P:p\mid n\}|,\qquad \mu=\sum_{p\mid P}\frac1p\to\infty,\qquad C=\sum_{p\mid P}\frac1{p^2}<1.$$

For uniform $n\bmod P$, $\mathbb EK=\mu$ and $\operatorname{Var}K\le\mu$. Chebyshev shows that at least half the residues have $K(n)\le t:=\mu+\sqrt{2\mu}$. Choose exactly $P/2$ of these residues as $S$, and set

$$\gamma=\frac1P\sum_{n\in S}\delta_{(n,n)}.$$

Its mass and marginals have the claimed properties.

Generate $n$ uniformly, then generate $m$ uniformly in the residue class $[n]_{(n,P)}$. The prime coordinates of $m$ are independent and

$$\Pr(p\mid m)=\frac2p-\frac1{p^2}.$$

Indeed a zero coordinate of $n$ forces a zero coordinate of $m$; otherwise the latter is uniform. Hence

$$\int f\,d\gamma =\Pr(n\in S,m\in S) \le\Pr(K(m)\le t) \le\frac{2\mu}{(\mu-C-\sqrt{2\mu})^2}\longrightarrow0.$$

The last inequality applies once $\mu-C-\sqrt{2\mu}>0$. A Chernoff bound with parameter $1/2$ gives the stronger bound

$$\int f\,d\gamma \le\exp\bigl(-(1-\log2)\mu+\sqrt{2\mu}\log2+C/2\bigr).$$

This is only a finite coupling obstruction. Such a coupling is not a construction of actual first-kill masses, and it does not account for the integer sizes needed to realize many logarithmic scales.

Primitive small-footprint families with a large atom kernel

There are finite primitive squarefree families $D$ with full multiples footprints $\beta_D\to0$, and cutoffs $X\to\infty$, for which

$$\frac1{\log X}\sum_{q\in D}\frac{\log\log q}{q}\longrightarrow\infty. \tag{4}$$

Thus the atom kernel arising when the coherent annulus estimate is summed over many bands cannot be controlled solely by primitivity and footprint.

Let $Y\to\infty$, put $L=\log\log Y$, and take the prime pool $\mathcal P=\{p:L^3<p\le Y\}$. Write

$$\lambda=\sum_{p\in\mathcal P}\frac1p =L-\log\log L+O(1),\qquad r=\left\lceil\lambda+\tfrac12\sqrt{\lambda\log\lambda}\right\rceil.$$

The reciprocal weight of all products of $r$ distinct pool primes is

$$S=\sum_{|A|=r}\frac1{\prod_{p\in A}p} =(1+o(1))\frac{\lambda^r}{r!} \asymp e^\lambda\lambda^{-5/8}. \tag{5}$$

Indeed, normalize the weights $1/p$ to sample $r$ primes independently. The collision probability is at most $\binom r2/(\lambda L^3)=O(L^{-2})$. Stirling's formula, with $r-\lambda=\tfrac12\sqrt{\lambda\log\lambda}+O(1)$, proves (5). The elementary Euler-product bounds already established in the wiki give $e^\lambda\asymp\log Y/\log L$.

For a Haar integer, let $N$ count the pool primes that divide it. Its mean is $\lambda$ and variance at most $\lambda$. Thus the full footprint of all these products is

$$\Pr(N\ge r)\le\frac{4}{\log\lambda}\longrightarrow0. \tag{6}$$

Restrict the products to $Y^a\le q\le Y^C$, where $a>0$ is a sufficiently small fixed constant and $C$ a sufficiently large fixed constant. This retains at least half of $S$. For the upper cutoff, the mean of $\log q$ under normalized reciprocal weights is at most

$$(1+o(1))\frac r\lambda \sum_{p\in\mathcal P}\frac{\log p}{p}\ll\log Y,$$

using the elementary Chebyshev prime-sum bound. Markov's inequality therefore discards at most $S/4$ when $C$ is large. For the lower cutoff, a product below $Y^a$ uses only primes below $Y^a$. Their reciprocal sum is at most $\lambda-c$, with arbitrarily large fixed $c$ available by taking $a$ small, since $\sum_{Y^a<p\le Y}1/p=\log(1/a)+O(1)$. Its total weight relative to $S$ is at most $(1+o(1))(1-c/\lambda)^r$, and is less than $1/4$ for an appropriate fixed $c$.

Let $D$ be the retained products and $X=Y^C$. The family is primitive because all its squarefree members have the same number $r$ of prime factors. Equation (6) still bounds its footprint. Since $\log\log q=L+O_{a,C}(1)$ throughout the retained range, (5) gives

$$\frac1{\log X}\sum_{q\in D}\frac{\log\log q}{q} \gg\frac{L^{3/8}}{\log L}\longrightarrow\infty,$$

proving (4).

Tautness does not imply the weighted reciprocal criterion

These finite blocks can be concatenated into an infinite primitive taut set with arbitrarily small full footprint and unbounded $W_P(X)/\log X$. Here taut means that deleting any one generator strictly decreases the full multiples-union measure.

Choose successive $Y_j$ so large that the prime pools $((\log\log Y_j)^3,Y_j]$ are disjoint and ordered, their trimmed modulus ranges are also ordered, and their footprints have sum less than a prescribed $\varepsilon<1$. Let $P$ be the union of the corresponding finite blocks. Disjoint prime supports and equal rank within each block make $P$ primitive. Its full footprint is at most $\varepsilon$. For any generator $q$ in block $j$, the event that exactly its $r_j$ pool primes divide a Haar integer has positive probability and makes $q$ the unique certificate in its block. Independently avoiding the other blocks has positive probability $\prod_{i\ne j}(1-\beta_i)>0$. Thus $q$ has a positive private Haar region, proving tautness.

At $X_j=Y_j^C$, the $j$th block alone gives

$$\frac{W_P(X_j)}{\log X_j} \ge\frac aC\sum_{q\in D_j}\frac1q\longrightarrow\infty$$

by (5). This refutes the proposed route from tautification to the weighted logarithmic criterion. It does not supply a coherent logarithmic-density anomaly.

Even diagonal profinite fibers need not return in Haar measure

For every $\varepsilon>0$ there is a compact $S\subseteq G=\widehat{\mathbb Z}$, with $\mu(S)>1-\varepsilon$, such that for Haar-almost every $r\in S$,

$$m_{rG}(S)=0. \tag{7}$$

Here $m_{rG}$ is normalized Haar measure on the closed subgroup $rG$, not a ratio of ambient Haar measures; that subgroup may have ambient measure zero.

Let $R,U$ be independent Haar points and put $L(y)=\sum_{p\le y}1/p$. Independence across primes gives

$$\frac{\#\{p\le y:p\mid R\}}{L(y)}\longrightarrow1 \quad\text{almost surely}.$$

For the product $RU$, the prime-divisibility indicators are also independent, now with probabilities $2/p-1/p^2$. Consequently their normalized count tends almost surely to $2$. Both assertions follow directly from Chebyshev and Borel--Cantelli along cutoffs with $L(y_k)=k^2+O(1)$: the variances are $O(L(y_k))$, so the relative error probabilities are summable. Monotonicity of the counts and $L(y_{k+1})/L(y_k)\to1$ interpolate between cutoffs.

Thus the law of $RU$ is singular to Haar measure. The Borel set where the displayed limit is $1$ has full Haar measure and zero product-law measure. Inner regularity supplies a compact subset $S$ of that set with $\mu(S)>1-\varepsilon$. Hence

$$0=\Pr(RU\in S)=\int_G m_{rG}(S)\,d\mu(r),$$

because multiplication by $r$ pushes Haar measure onto Haar measure of its compact image $rG$. Nonnegativity proves (7).

For example, the diagonal measure obtained by sending $\mathbf1_S\mu$ under $r\mapsto(q,n)=(r,r)$ has two marginals bounded by Haar measure and lies on $n\in S$. Nevertheless almost every corresponding affine ideal $n+qG=rG$ has zero normalized Haar intersection with $S$. This defeats a proposed positive-fiber return argument based only on those marginals, even for $q=n$. It constructs no positive integer private endpoints: a typical profinite $r$ has infinitely many prime divisors, and no comparison between a numerical modulus and an integer witness is supplied.