Profinite residue endpoints

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The results here concern bounded-quotient first kills, not the full delayed union. They do not resolve Problem 25. They prove logarithmic nullity whenever the normalized residues have countable profinite closure, and give a necessary distributional condition in general.

Relative compression: proved

Let $S\subseteq\widehat{\mathbb Z}$ be $P$-periodic, and suppose a finite or countable family $C_i=[a_i]_{q_i}$ satisfies

$$a_i\equiv c\pmod{(q_i,P)} \quad\text{for every }i,$$

where $c$ is one integer. Then

$$\mu\left(S\cap\bigcup_i[c]_{q_i}\right) \le \mu\left(S\cap\bigcup_i C_i\right). \tag{1}$$

Distinctness of the moduli is not needed for this lemma.

For a finite family work modulo $L=\operatorname{lcm}(P,q_i)$ and use CRT. Fix a prime $p\mid L$, all other prime coordinates, and the low digits $x\pmod{p^v}$, where $v=v_p(P)$. A cylinder with $e=v_p(q_i)\le v$ is either empty or the entire remaining fiber; replacing its center by $c$ at this prime changes neither possibility. When $e>v$, the cylinder meets only the low fiber $c\pmod{p^v}$. Inside that fiber its restriction is a $p$-adic ball. Moving all the remaining centers to $c$ makes these balls nested, so their union has measure equal to the largest ball and cannot increase. An entire-fiber restriction, if present, also causes no problem. Since $S$ is constant on each fixed $P$-residue, the same comparison holds after intersection with $S$. Integrate and repeat over all prime coordinates. This proves (1) for finite families; monotone convergence proves the countable case.

Multiples unions intersected with a periodic set

For arbitrary $D\subseteq\mathbb N$ and a periodic set $R$, the set $M(D)\cap R$ has logarithmic density

$$\operatorname{ld}(M(D)\cap R)=\mu(M(D)\cap R). \tag{2}$$

Here $M(D)$ denotes the positive multiples union or its Haar version, according to the ambient space.

Indeed, the exact multiples theorem gives $\operatorname{ld}(M(D))=\beta:=\mu(M(D))$. For finite $D_j\uparrow D$, put $\beta_j=\mu(M(D_j))$. Subtracting the finite periodic harmonic sum gives

$$\operatorname{ld}(M(D)\setminus M(D_j))=\beta-\beta_j\longrightarrow0.$$

The periodic sets $M(D_j)\cap R$ thus approximate $M(D)\cap R$ in upper logarithmic density. Their densities converge to the right side of (2) by Haar monotone convergence. This proves (2); no natural-density assertion for the infinite multiples union is being made.

A residue-cylinder bound for first endpoints

Use strictly increasing moduli $q_i$, normalized residues $0\le a_i<q_i$, and notation

$$C_i=[a_i]_{q_i},\quad U=\bigcup_i C_i,\quad B_i=\{a_i+tq_i:t\ge1\},\quad E_i=B_i\setminus\bigcup_{h<i}B_h.$$

Fix $K\ge1$, write $m=\lfloor K\rfloor$, and define

$$T_K=\bigcup_i(E_i\cap[q_i,Kq_i]).$$

Every point in $T_K$ has a unique witnessing index $i(n)$ and an integer quotient $1\le t=(n-a_{i(n)})/q_{i(n)}\le m$.

For a prefix $j$, put

$$U_j=\bigcup_{i\le j}C_i,\quad \tau_j=\mu(U\setminus U_j),\quad P_j=\operatorname{lcm}(q_1,\ldots,q_j).$$

For every residue $c\pmod{P_j}$,

$$\overline{\operatorname{ld}} \{n\in T_K:a_{i(n)}\equiv c\pmod{P_j}\} \le m\tau_j. \tag{3}$$

The coefficient does not depend on the old period.

To prove this, discard the finitely many points with $i(n)\le j$ and write $S=\widehat{\mathbb Z}\setminus U_j$ and $R=S-c$. Every remaining first kill lies in $S$: at $n\ge q_{i(n)}>q_j$, all old congruences are active. For each $t\le m$, let $D_t$ consist of the moduli of points $n=a_i+tq_i$ under consideration. Then $tq_i\in R$ and $n\ge tq_i$. Distinctness makes $q_i\mapsto tq_i$ injective, so

$$H_{\{n\in T_K:i(n)>j,\ a_{i(n)}\equiv c\ (P_j)\}}(X) \le\sum_{t=1}^m H_{M(tD_t)\cap R}(X). \tag{4}$$

By (2), each right-hand term has logarithmic density equal to its Haar measure. Since $M(tD_t)\subseteq M(D_t)$, translation and (1) give

$$\begin{aligned} \mu(M(tD_t)\cap R) &\le\mu((c+M(D_t))\cap S)\\ &\le\mu\left(S\cap\bigcup_{q_i\in D_t}C_i\right) \le\tau_j. \end{aligned}$$

Here $a_i\equiv c\pmod{P_j}$ is stronger than the compatibility hypothesis of (1). Summing (4) proves (3), with the original activation rule unchanged.

Diffuse limiting residues and a logarithmic-nullity criterion

For $X\ge2$ define a finite measure on $G=\widehat{\mathbb Z}$ by

$$\nu_X=\frac1{\log X}\sum_{\substack{n\le X\\n\in T_K}} \frac1n\,\delta_{a_{i(n)}}. \tag{5}$$

Every weak limit along $X\to\infty$ of these measures is nonatomic. Their masses are bounded and $G$ is compact metrizable. For a weak limit $\nu$, residue-cylinder indicators are continuous, so (3) gives

$$\nu([c]_{P_j})\le m\tau_j\qquad\text{for every }j,c.$$

For any $\alpha\in G$,

$$\nu(\{\alpha\})\le\nu([\alpha]_{P_j})\le m\tau_j\longrightarrow0.$$

The periods $P_j$ need not exhaust all finite moduli for this deduction.

Corollary. If the normalized residues $\{a_i\}$ have countable closure in $\widehat{\mathbb Z}$, then

$$\operatorname{ld}(T_K)=0\qquad\text{for every fixed }K. \tag{6}$$

In particular, this holds if $a_i$ converges profinitely.

Every measure (5), and hence every weak limit, is supported on that closed countable set. A nonatomic measure supported there is zero. If the masses $H_{T_K}(X)/\log X$ did not tend to zero, compactness would give a weak limit with positive mass, a contradiction. A convergent sequence together with its limit has countable closure, proving the last assertion.

For arbitrary residues the closure may be uncountable, and diffuse limits are possible in principle. Nothing here controls that case or first kills with unbounded quotient. In particular, (6) alone does not give logarithmic density for the original survivor set.

The binary-colored private-witness construction shows a further limitation of purely profinite endpoint arguments. Even locally $0/1$-colored idempotent congruences can have private profinite witnesses, a modulus set of natural density $1/4$, and positive survivor density $1/2$. Its actual delayed union does have natural density. Short positive integer witnesses, not merely profinite privacy or a bounded local color alphabet, remain essential.