Irrational slope endpoints

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These results concern the bounded-quotient first-kill problem, not the full delayed union. The positive result controls conditioning on an old period without a constant growing with that period. The original badly approximable theorem below extends to every Brjuno slope.

A different curvature argument proves natural endpoint nullity for $\lfloor\theta q+q^\alpha\rfloor$, $0<\alpha<1$, for every slope $\theta$, including Liouville slopes. It changes the residue profile; it does not remove the hypothesis from the purely linear theorem below or control unbounded quotients.

Badly approximable slopes: proved natural nullity

Fix $0<\theta<1$ with

$$c_\theta:=\inf_{k\ge1}k\|k\theta\|>0,$$

and put $a_q=\lfloor\theta q\rfloor$, $C_q=[a_q]_q$. Let $D$ be a set of distinct moduli and fix an integer $\ell\ge1$. Suppose

$$r_q=\ell q+a_q\notin C_p \qquad(p,q\in D,\ p<q). \tag{1}$$

Then both $D$ and $\{r_q:q\in D\}$ have natural density zero. Full privacy is not required.

Consequently, in any sieve with these residues, the union of first-kill points with $q\le n\le Kq$ has natural density zero for every fixed $K$. Partition those points by the finitely many possible quotient indices $\ell=(n-a_q)/q$ and apply the theorem to each selected modulus set.

A uniform dyadic second moment

For $D_Q\subseteq[Q,2Q)\cap\mathbb N$, let $Y_Q=\sum_{q\in D_Q}\mathbf1_{C_q}$ on Haar space. Then

$$\mathbb EY_Q^2\le C_\theta, \qquad C_\theta=10+16/c_\theta \tag{2}$$

is one possible uniform bound.

Indeed, for compatible distinct $q,r$, write $g=(q,r)$ and $k=|q/g-r/g|$. Compatibility implies

$$1\le k\le Q/g,\qquad \|k\theta\|<1/g,$$

because $\lfloor\theta q\rfloor-\lfloor\theta r\rfloor$ is divisible by $g$ and differs from $\theta(q-r)$ by less than one. Among $1\le k\le T$, the fractional parts $k\theta$ have circular separation at least $c_\theta/T$. Hence at most $2+2Q/(c_\theta g^2)$ possible $k\le Q/g$ qualify. Each gives at most $4Q/g$ ordered pairs $(q,r)$ in the block. Since $g<Q$ for distinct pairs,

$$\sum_{\substack{q,r\in[Q,2Q)\\q\ne r,\ C_q\cap C_r\ne\varnothing}} (q,r) \le\sum_{g<Q}4Q\left(2+\frac{2Q}{c_\theta g^2}\right) \le(8+16/c_\theta)Q^2.$$

Divide by $qr\ge Q^2$ and add the diagonal sum $\sum1/q\le1$. This proves (2).

A reciprocal cost that does not grow with the old period

Fix a finite initial part $F\subset D$. Let

$$U=\bigcup_{q\in D}C_q,\quad U_F=\bigcup_{q\in F}C_q, \quad \tau_F=\mu(U\setminus U_F),$$

and let $S\subseteq\mathbb Z/P\mathbb Z$ be the survivor residues of $U_F$ for a common period $P$. If $S$ is empty, (1) permits no later selected endpoint, so there is nothing to prove.

For $q$ beyond $F$, define $d=(q,P)$ and

$$f(q)=\frac{|S\cap[a_q]_d|}{P/d}.$$

The endpoint condition gives $f(q)>0$, and CRT gives $\mu(C_q\setminus U_F)=f(q)/q$.

For fixed residues $(t,b)=(q\bmod P,a_q\bmod P)$, admissibility means $b+\ell t\in S$. Writing $d=(t,P)$, let $f(t,b)$ be the same survivor fraction. Then

$$\sum_{\substack{b\bmod P\\b+\ell t\in S}}\frac1{f(t,b)} =\frac Pd\,\#\{\text{nonempty }S\text{-fibers modulo }d\} \le P. \tag{3}$$

The equality holds because $d\mid t$: in each nonempty residue fiber, there are exactly $|S\cap[b]_d|$ admissible values of $b$, each with reciprocal cost $(P/d)/|S\cap[b]_d|$.

For fixed $P$, the pairs $(q\bmod P,\lfloor\theta q\rfloor\bmod P)$ have frequency $1/P^2$. To see this, write $q=t+Pm$: the second residue is determined by which of the $P$ equal intervals contains $\{\theta m+\theta t/P\}$. Irrational rotation is equidistributed, as follows from the geometric sums of its nonconstant Fourier characters. Consequently (3), summed over $t$, gives

$$\sum_{\substack{Q\le q<2Q\\a_q+\ell q\bmod P\in S}} \frac1{f(q)}\le Q+o_{P,\theta}(Q). \tag{4}$$

The period is held fixed when taking this limit.

Let $m_Q=|D\cap[Q,2Q)|$. Cauchy--Schwarz and (4) imply

$$\sum_{q\in D\cap[Q,2Q)}f(q) \ge\frac{m_Q^2}{Q+o_{P,\theta}(Q)}.$$

The nonnegative function $Z_Q=\sum_{q\in D\cap[Q,2Q)}\mathbf1_{C_q\setminus U_F}$ has support of measure at most $\tau_F$ and second moment at most $C_\theta$. Thus

$$\frac{m_Q^2}{2Q(Q+o_{P,\theta}(Q))} \le\mathbb EZ_Q\le\sqrt{C_\theta\tau_F},$$

and hence

$$\limsup_{Q\to\infty}\frac{m_Q}{Q} \le(4C_\theta\tau_F)^{1/4}.$$

Now exhaust $D$ by finite prefixes, so $\tau_F\to0$. Dyadic block densities tend to zero, proving $d(D)=0$. Since $r_q\ge q$, the endpoint set also has natural density zero.

Extension to Brjuno slopes: proved

Let $b_n$ be the continued-fraction denominators of an irrational $0<\theta<1$. The same natural-nullity conclusion holds under the weaker hypothesis

$$\mathcal B(\theta):=\sum_{n\ge0}\frac{\log b_{n+1}}{b_n}<\infty. \tag{B}$$

This is the Brjuno condition. In particular, it holds for every irrational of finite irrationality exponent.

The only new ingredient needed in the preceding proof is

$$\mathbb E\left(\sum_{Q\le q<2Q} \mathbf1_{[\lfloor\theta q\rfloor]_q}\right)^2 \le C(1+\mathcal B(\theta)), \tag{6}$$

with an absolute constant $C$. Subfamilies obey the same bound by pointwise domination. We prove (6) next.

For compatible distinct $q,r\in[Q,2Q)$, write $g=(q,r)$ and $k=|q-r|/g$. As above,

$$k<Q/g,\qquad \|k\theta\|<1/g,$$

and at most $4Q/g$ ordered pairs correspond to fixed $g,k$. Summing their contributions $g/(qr)\le g/Q^2$ and then interchanging $g,k$ gives

$$\mathbb EY_Q^2\le1+4S_\theta(Q),\qquad S_\theta(Q)=\sum_{k\le Q} \min\left(\frac1k,\frac1{Q\|k\theta\|}\right). \tag{7}$$

Partition $k\le Q$ into dyadic blocks $T\le k<2T$, truncated at $Q$, and take the largest convergent denominator $b=b_n\le T$. Fractional parts in any block of at most $b$ consecutive integers are circularly separated by more than $1/(2b)$. Here is a short verification of the continued-fraction fact used in that assertion. Consecutive convergent vectors $(b_{n-1},p_{n-1})$ and $(b_n,p_n)$ form an integer basis, and their errors $\delta_{n-1}=b_{n-1}\theta-p_{n-1}$ and $\delta_n=b_n\theta-p_n$ have opposite signs. Write any integer pair $(h,m)$ in this basis. If $0<h<b_n$, its coefficients cannot both be nonnegative with a positive coefficient at $b_n$, nor both be nonpositive. Opposite nonzero signs make the two error terms have the same sign, with absolute sum at least $|\delta_{n-1}|$. The remaining case is a positive multiple of the previous vector, which has the same lower bound. Thus $\|h\theta\|\ge|\delta_{n-1}|$ for $0<h<b_n$. Writing the continued-fraction tail after the $n$th convergent as $u>1$ gives

$$|\delta_{n-1}|=\frac{u}{b_nu+b_{n-1}} >\frac1{b_n+b_{n-1}}\ge\frac1{2b_n}.$$

The case $b=1$ is vacuous. Initial repeated denominator $1$ may simply be omitted when selecting the largest denominator.

Partition a dyadic block into at most $2T/b$ consecutive blocks of at most $b$ integers. The separation just proved implies, for $0<u\le1/2$,

$$F_T(u):=\#\{T\le k<\min(2T,Q+1):\|k\theta\|\le u\} \le8Tu+\frac{2T}{b}. \tag{8}$$

For $T\le Q/2$, integrate the distribution function of $\min(1/T,1/(Q\|k\theta\|))$ to obtain

$$\begin{aligned} S_T &\le\frac{2T}{Q}+ \frac1Q\int_{T/Q}^{1/2}\frac{F_T(u)}{u^2}\,du\\ &\le\frac{2T}{Q}+ \frac{8T}{Q}\log(Q/T)+\frac2b. \end{aligned} \tag{9}$$

For $T>Q/2$, the same upper bound follows from $S_T\le1$. The first two terms in (9) sum to an absolute constant over dyadic $T$. A denominator $b_n$ is selected only for $b_n\le T<b_{n+1}$, so

$$\sum_{T\text{ dyadic}}\frac1{b(T)} \le\sum_n\frac{1+\log_2(b_{n+1}/b_n)}{b_n} \ll1+\mathcal B(\theta).$$

The harmless sum $\sum1/b_n$ is absolutely bounded, because $b_{n+2}\ge2b_n$. Equations (7)--(9) prove (6).

The preceding period-independent reciprocal-cost argument uses only irrational equidistribution apart from the second moment. Replacing $C_\theta$ by $C(1+\mathcal B(\theta))$ in that argument therefore gives

$$\limsup_{Q\to\infty}\frac{|D\cap[Q,2Q)|}{Q} \le\bigl(4C(1+\mathcal B(\theta))\tau_F\bigr)^{1/4}.$$

Exhausting $D$ by finite prefixes proves natural nullity. The same proof allows an added fixed integer $c$ in $a_q=\lfloor\theta q\rfloor+c$: compatibility differences are unchanged and the residue-pair distribution is merely translated. These are normalized residues for all sufficiently large $q$; discarding the remaining finite set does not affect density. Finitely many such perturbations or slopes satisfying (B) are handled by partitioning the selected endpoints into finitely many subfamilies.

To check the finite-irrationality-exponent assertion, an inequality $|\theta-a/b|\ge c b^{-M}$ for sufficiently large denominators, together with the convergent estimate $|\theta-p_n/b_n|<1/(b_nb_{n+1})$, gives $b_{n+1}\le c^{-1}b_n^{M-1}$. Thus $\log b_{n+1}=O(\log b_n)$, and the at-least-Fibonacci growth of $b_n$ makes (B) converge.

Why the remaining irrational logarithmic case is not settled

For a convergent $A/b$, its approximation has error less than $1/(b b_{\rm next})$. For $q<b_{\rm next}$ this gives the same floor as $Aq/b$, except that $\theta<A/b$ subtracts one when $b\mid q$. This is an exact reduction of the long approximation epoch to finitely many rational patterns. It does not give a uniform estimate independent of $b$. The verified rational-pattern proofs have denominator-dependent losses, whereas (7) can already grow with $(\log Q)/b$ inside such an epoch. No estimate bridging those regimes has been proved. In particular, the Brjuno theorem does not establish logarithmic endpoint nullity for every irrational slope.

Irrationality alone does not give natural nullity

There is a Liouville number $\theta\in(1/4,1/2)$ and a globally private family

$$r_q=q+\lfloor\theta q\rfloor,\qquad q\le r_q<2q,$$

whose endpoint set has upper natural density at least $1/16$ but logarithmic density zero. Its full footprint can be arbitrarily small. This does not contradict logarithmic endpoint nullity.

For a fixed rational slope $A/b$, its full dyadic block union obeys

$$\mu\left(\bigcup_{Q\le q<2Q}[\lfloor Aq/b\rfloor]_q\right) \le b^2\mu(M([Q,2Q)))\longrightarrow0. \tag{5}$$

Indeed every hit satisfies $q\mid bn+t$ for some $0\le t<b$. The Haar measure of each affine preimage is at most $b$ times the multiples-union measure, because multiplication by $b$ maps Haar probability onto normalized Haar measure on $b\widehat{\mathbb Z}$. The multiples-window limit is already proved in common-shift blocks.

Fix $0<\varepsilon<1/4$. Construct increasing rational approximants $\theta_j=A_j/b_j$ inside nested small intervals in $(1/4,1/2)$, with $b_j\to\infty$. When the earlier selected cylinders have common period $P$, require $(A_j+b_j,P)=1$. Such rationals are dense: take $b_j$ divisible by $P$ and $A_j\equiv1\pmod P$.

Choose a sufficiently large dyadic $Q_j$ that the block footprint in (5) is below $\varepsilon2^{-j}$, that $Q_j\gg b_jP$, and that $Q_j/4-1$ exceeds every earlier endpoint. Among $Q_j\le q<5Q_j/4$, retain endpoints $q+\lfloor\theta_jq\rfloor$ avoiding the earlier full union. Their residues modulo $P$ are uniform over a period $b_jP$: increasing $q$ by $b_j$ increases the endpoint by $A_j+b_j$, a unit modulo $P$. Since the earlier union has measure at most $\varepsilon$, at least $Q_j/8$ endpoints can be retained for sufficiently large $Q_j$.

Their diameter is less than $Q_j$, so their classes are private within the block. Every new least positive representative is greater than all earlier endpoints, establishing reverse-direction privacy as well.

After choosing this block, take the next closed approximation interval strictly to the right of $\theta_j$ and inside

$$\left(\theta_j,\theta_j+ \min\{1/(2b_jQ_j),b_j^{-j}\}\right),$$

also inside the preceding approximation interval. The limit $\theta$ therefore satisfies the indicated one-sided bounds for every $j$. For $q<2Q_j$, changing the slope by less than $1/(2b_jQ_j)$ cannot change the floor: the next rational integer boundary is at distance at least $1/b_j$. The strict one-sided approximations with unbounded denominators make the limit irrational and Liouville.

All constructed cylinders and endpoints thus persist for this single $\theta$. Each block has at least $Q_j/8$ endpoints below $2Q_j$, so $\overline d(E)\ge1/16$. Its block footprints are summable; the summable-footprint endpoint lemma gives $\operatorname{ld}(E)=0$.

The positive theorem's Diophantine hypothesis therefore cannot simply be replaced by irrationality if its natural-density conclusion is retained. Neither construction resolves logarithmic nullity for general private endpoints or controls unbounded-quotient first kills.