Erdős Problem #38

Erdős Problem #38 --- Round 4 (continuation & gap-closure)

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction.

Round 3 pushed the one-block periodic obstruction as far as it naturally goes, obtaining the necessary bound \[ f(\alpha)\le \min(\alpha,1-\alpha) \] for rational $\alpha$, together with central-interval constraints on any candidate set $B$. The most promising remaining gap is the classical "random-set barrier": Erdős observed that even for $B=\mathbb N$ one should not hope for an improvement larger than $\alpha(1-\alpha)$. Round 3 did not make this rigorous.

The goal of Round 4 is therefore to replace the highly structured block test sets by pseudorandom periodic test sets and prove a strictly sharper universal obstruction: \[ f(\alpha)\le \alpha(1-\alpha) \qquad\text{for every rational }\alpha\in(0,1). \] This is a genuine strengthening of the Round 3 restriction, since \[ \alpha(1-\alpha)\le \min(\alpha,1-\alpha) \] with strict inequality for every $\alpha\in(0,1)\setminus\{1/2\}$.

2) ROUND-3 FOUNDATION USED

I use the following specific Round 3 ingredients.

• Round 3 formal restatement and notation for Erdős Problem #38.
• Round 3 periodic-test-set viewpoint: the property can be challenged by constructing a single set $A$ of prescribed Schnirelmann density and evaluating the inequality at one carefully chosen $N$.
• Round 3 obstruction baseline: for rational $\alpha$, one already has the weaker necessary bound \[ f(\alpha)\le \min(\alpha,1-\alpha). \] The new work in this round is to sharpen this to $f(\alpha)\le \alpha(1-\alpha)$.

I will not re-prove the Round 3 interval lemmas; instead I replace that entire obstruction family by a different periodic construction.

3) NEW INSIGHT / TOOL (ROUND-4)

The new ingredient is a rotation-and-randomisation principle for periodic sets.

• First, any binary word of length $Q$ with exactly $\alpha Q$ ones can be rotated so that its periodic extension has exact Schnirelmann density $\alpha$. This is a cycle-lemma type observation obtained by rotating at a minimum of the discrepancy function.
• Second, a random binary word of density $\alpha$ has, with positive probability, near-random pair correlations for every nonzero cyclic shift simultaneously. After conditioning on having exactly $\alpha Q$ ones, this gives a deterministic period whose cyclic self-intersections all satisfy \[ |R\cap(R+t)|\ge (\alpha^2-\varepsilon)Q \qquad (t\neq 0). \]
• Combining the two facts yields a periodic set $A$ of exact Schnirelmann density $\alpha$ for which every shift $b$ satisfies \[ |(A\cup(A+b))\cap[1,Q]|\le (\alpha+\alpha(1-\alpha)+\varepsilon)Q. \] This turns Erdős' random heuristic into a rigorous rational-density obstruction.

4) ATTACK PLAN (ROUND-4)

The exact gap left after Round 3 is that all previous obstructions came from highly structured interval-type periods, so they only forced the coarse upper bound $f(\alpha)\le \min(\alpha,1-\alpha)$. To go beyond that I will:

1. prove a rotation lemma showing that any cyclic word with density $\alpha$ can be turned into an infinite periodic set with Schnirelmann density exactly $\alpha$;
2. prove, by the probabilistic method, that for every rational $\alpha$ and every $\varepsilon>0$ there exists a period whose nonzero cyclic shifts all have self-intersection at least $(\alpha^2-\varepsilon)Q$;
3. use that period as a test set at $N=Q$ to show that no shift can improve the density by more than $\alpha(1-\alpha)+\varepsilon$.

This directly overcomes the main Round 3 obstacle: instead of trying to optimise within the interval family, it switches to a family whose correlations are asymptotically as small as the random model allows.

5) WORK (ROUND-4)

5.1 Rotation lemma: turning a cyclic word into a Schnirelmann-extremal periodic set

Let $Q\ge 2$, let $0<L<Q$, and put $\alpha=L/Q$. Let $w=(w_1,\dots,w_Q)\in\{0,1\}^Q$ with \[ \sum_{i=1}^Q w_i=L. \] For $0\le n\le Q$ define the discrepancy sums \[ D_n:=\sum_{i=1}^n (w_i-\alpha), \qquad D_0:=0. \] Since $D_Q=L-\alpha Q=0$, the finite set $\{D_0,D_1,\dots,D_{Q-1}\}$ has a minimum. Choose $m\in\{0,1,\dots,Q-1\}$ such that \[ D_m=\min_{0\le j\le Q-1} D_j. \] Now rotate the word by $m$ places and define \[ w'_r := w_{m+r} \qquad (1\le r\le Q), \] with indices understood modulo $Q$.

Lemma 5.1 (rotation lemma).

For every $1\le r\le Q$ one has \[ \sum_{i=1}^r w'_i \ge \alpha r. \] Consequently, if $A\subseteq\mathbb N$ is the infinite periodic set obtained by repeating $w'$ with period $Q$, then \[ d_s(A)=\alpha. \]

Proof. Fix $1\le r\le Q$. If $m+r\le Q$, then \[ \sum_{i=1}^r w'_i - \alpha r = \sum_{i=m+1}^{m+r}(w_i-\alpha) = D_{m+r}-D_m\ge 0 \] because $D_m$ is minimal. If $m+r>Q$, then using $D_Q=0$ we get \[ \sum_{i=1}^r w'_i - \alpha r = \sum_{i=m+1}^{Q}(w_i-\alpha)+\sum_{i=1}^{m+r-Q}(w_i-\alpha) = (D_Q-D_m)+D_{m+r-Q} = D_{m+r-Q}-D_m\ge 0. \] Thus every prefix of $w'$ has at least $\alpha$-density.

Let $c_r:=\sum_{i=1}^r w'_i$ for $0\le r\le Q$ (with $c_0=0$). Then $c_r\ge \alpha r$ for all $r$, and $c_Q=L=\alpha Q$. If $N=kQ+r$ with $k\ge 0$ and $0\le r<Q$, then the periodic extension $A$ satisfies \[ |A\cap[1,N]| = kL+c_r \ge k\alpha Q + \alpha r = \alpha N. \] Hence $d_s(A)\ge \alpha$. On the other hand, for every $k\ge 1$, \[ \frac{|A\cap[1,kQ]|}{kQ} = \frac{kL}{kQ}=\alpha, \] so $d_s(A)\le \alpha$. Therefore $d_s(A)=\alpha$. $\square$

5.2 Pseudorandom periods with exact density

Fix a rational number $\alpha\in(0,1)$ and $\varepsilon>0$. Write $\alpha=a/q_0$ in lowest terms. Let $Q$ be a large multiple of $q_0$, so that $\alpha Q\in\mathbb N$.

Consider independent Bernoulli random variables $(\xi_x)_{x\in\mathbb Z/Q\mathbb Z}$ with \[ \mathbb P(\xi_x=1)=\alpha, \qquad \mathbb P(\xi_x=0)=1-\alpha. \] Let \[ R:=\{x\in\mathbb Z/Q\mathbb Z: \xi_x=1\}. \] For each nonzero $t\in\mathbb Z/Q\mathbb Z$, define \[ X_t:=|R\cap(R+t)| = \sum_{x\in\mathbb Z/Q\mathbb Z} \xi_x\xi_{x+t}. \] Also set \[ Y:=|R|=\sum_{x\in\mathbb Z/Q\mathbb Z}\xi_x. \]

Lemma 5.2 (uniform correlation lower bound on one random period).

For each fixed nonzero $t\in\mathbb Z/Q\mathbb Z$ one has \[ \mathbb E X_t = \alpha^2 Q, \] and \[ \mathbb P\left(X_t< (\alpha^2-\varepsilon)Q\right) \le \exp\left(-\frac{\varepsilon^2Q}{2}\right). \] Hence \[ \mathbb P\left(\forall t\neq 0,\ X_t\ge (\alpha^2-\varepsilon)Q\right) \ge 1-(Q-1)\exp\left(-\frac{\varepsilon^2Q}{2}\right). \]

Proof. For $t\neq 0$, the variables $\xi_x$ and $\xi_{x+t}$ are independent, so \[ \mathbb E(\xi_x\xi_{x+t})=\alpha^2. \] Summing over $x\in\mathbb Z/Q\mathbb Z$ gives $\mathbb E X_t=\alpha^2Q$.

Now view $X_t$ as a function of the independent family $(\xi_x)_{x\in\mathbb Z/Q\mathbb Z}$. Changing a single coordinate $\xi_y$ can affect only the two summands \[ \xi_y\xi_{y+t} \qquad\text{and}\qquad \xi_{y-t}\xi_y, \] so the value of $X_t$ changes by at most $2$. By the bounded-differences inequality, \[ \mathbb P\left(X_t\le \mathbb EX_t-u\right) \le \exp\left(-\frac{2u^2}{\sum_{x\in\mathbb Z/Q\mathbb Z} 2^2}\right) =\exp\left(-\frac{u^2}{2Q}\right). \] Taking $u=\varepsilon Q$ gives the displayed tail bound. The final estimate follows by the union bound over the $Q-1$ nonzero shifts. $\square$

Lemma 5.3 (forcing the random period onto the exact rational slice).

There exists $Q_1=Q_1(\alpha,\varepsilon)$ such that for every multiple $Q\ge Q_1$ of $q_0$ there is a deterministic set \[ R\subseteq \mathbb Z/Q\mathbb Z \qquad\text{with}\qquad |R|=\alpha Q \] and \[ |R\cap(R+t)|\ge (\alpha^2-\varepsilon)Q \qquad\text{for every }t\neq 0. \]

Proof. The event $\{Y=\alpha Q\}$ has probability \[ \mathbb P(Y=\alpha Q) = \binom{Q}{\alpha Q}\alpha^{\alpha Q}(1-\alpha)^{(1-\alpha)Q}. \] By Stirling's formula there exists a constant $c_\alpha>0$ such that, for all sufficiently large multiples $Q$ of $q_0$, \[ \mathbb P(Y=\alpha Q)\ge c_\alpha Q^{-1/2}. \] On the other hand, Lemma 5.2 gives \[ \mathbb P\left(\exists t\neq 0:\ X_t< (\alpha^2-\varepsilon)Q\right) \le (Q-1)\exp\left(-\frac{\varepsilon^2Q}{2}\right). \] Since $Qe^{-\varepsilon^2Q/2}=o(Q^{-1/2})$, for all sufficiently large $Q$ we have \[ \mathbb P(Y=\alpha Q) > \mathbb P\left(\exists t\neq 0:\ X_t< (\alpha^2-\varepsilon)Q\right). \] Therefore \[ \mathbb P\left(Y=\alpha Q\ \text{and}\ \forall t\neq 0,\ X_t\ge (\alpha^2-\varepsilon)Q\right)>0. \] So some deterministic realisation of $R$ has all the required properties. $\square$

5.3 Sharpened obstruction on the admissible size of $f(\alpha)$

We now turn the pseudorandom period into a counterexample to any putative inequality with gain exceeding $\alpha(1-\alpha)$.

Theorem 5.4 (random-model ceiling, rigorously proved for rational densities).

Assume that $B\subseteq\mathbb N$ satisfies the Erdős Problem #38 property with some function $f:(0,1)\to(0,\infty)$. Then for every rational $\alpha\in(0,1)$ one has \[ f(\alpha)\le \alpha(1-\alpha). \]

Proof. Fix rational $\alpha\in(0,1)$ and suppose, for contradiction, that \[ f(\alpha)>\alpha(1-\alpha). \] Choose \[ \varepsilon:=\frac{f(\alpha)-\alpha(1-\alpha)}{3}>0. \] By Lemma 5.3, for some sufficiently large multiple $Q$ of the denominator of $\alpha$ there exists a set \[ R\subseteq \mathbb Z/Q\mathbb Z \qquad\text{with}\qquad |R|=\alpha Q \] such that \[ |R\cap(R+t)|\ge (\alpha^2-\varepsilon)Q \qquad (t\neq 0). \] Apply Lemma 5.1 to a binary word representing $R$. After a cyclic rotation, we obtain an infinite periodic set $A\subseteq\mathbb N$ with \[ d_s(A)=\alpha \] and whose first period has residue set equal to a translate of $R$. Since cyclic translation preserves all intersection numbers $|R\cap(R+t)|$, we may continue to denote the first-period residue set by $R$.

Now set $N:=Q$. Fix any positive integer $b$.

If $b\ge Q$, then $(A+b)\cap[1,Q]=\varnothing$, so \[ |(A\cup(A+b))\cap[1,Q]| = |A\cap[1,Q]|=\alpha Q. \]

If $1\le b\le Q-1$, then shifting inside the first period can only lose wrapped points, so \[ |(A\cup(A+b))\cap[1,Q]| \le |R\cup(R+b)|, \] where $R+b$ denotes cyclic translation in $\mathbb Z/Q\mathbb Z$. Hence \[ |(A\cup(A+b))\cap[1,Q]| \le 2|R|-|R\cap(R+b)| \le 2\alpha Q-(\alpha^2-\varepsilon)Q = \left(\alpha+\alpha(1-\alpha)+\varepsilon\right)Q. \] Therefore for every $b\in\mathbb N$, \[ |(A\cup(A+b))\cap[1,Q]| \le \left(\alpha+\alpha(1-\alpha)+\varepsilon\right)Q. \] By the choice of $\varepsilon$, \[ \alpha+\alpha(1-\alpha)+\varepsilon < \alpha + f(\alpha). \] So for this set $A$ and this $N=Q$, no $b\in B$ can satisfy \[ |(A\cup(A+b))\cap[1,Q]|\ge (\alpha+f(\alpha))Q, \] contradicting the defining property of $B$.

This contradiction proves \[ f(\alpha)\le \alpha(1-\alpha) \qquad\text{for every rational }\alpha\in(0,1). \] $\square$

5.4 Consequences and comparison with Round 3

Theorem 5.4 strictly strengthens the Round 3 rational obstruction.

Corollary 5.5.

Under the Erdős Problem #38 hypothesis, for every rational $\alpha\in(0,1)$, \[ 0<f(\alpha)\le \alpha(1-\alpha)\le \min(\alpha,1-\alpha). \] In particular, \[ f(1/2)\le 1/4. \]

Proof. The first inequality is the problem hypothesis, the second is Theorem 5.4, and the third is elementary. $\square$

This also rules out an entire class of proof strategies: one cannot hope to settle the problem by proving the stronger statement with some function $f$ that exceeds the random-set barrier $\alpha(1-\alpha)$ at even a single rational density.

6) ADVERSARIAL VERIFICATION

• Does the rotation lemma really force positive Schnirelmann density, not just asymptotic density? Yes. The key point is that after rotating at a discrepancy minimum, every prefix of the first period has at least $\alpha$-density. Repeating the period then preserves this bound on every prefix of the infinite set. In particular the first entry of the rotated word must be $1$, so the "missing $1$" pathology from Round 2 cannot occur.

• Could the probabilistic construction fail after conditioning on the exact size $|R|=\alpha Q$? No conditioning argument is needed. The proof only uses that the event "$|R|=\alpha Q$" has probability $\gg Q^{-1/2}$ while the bad-correlation event has exponentially small probability. The difference of these two probabilities is positive for all sufficiently large $Q$.

• Is the bounded-differences estimate valid for $X_t$? Yes. For fixed $t\neq 0$, changing one Bernoulli coordinate $\xi_y$ changes only the two summands with indices $x=y$ and $x=y-t$. Hence the Lipschitz constant is exactly $2$.

• Could boundary effects at $N=Q$ invalidate the cyclic correlation bound? No. Boundary effects only remove wrapped points from $(A+b)\cap[1,Q]$. Therefore the actual first-period union is bounded above by the cyclic union $|R\cup(R+b)|$; this is precisely the inequality needed for the contradiction.

• Interaction with Round 3. There is no conflict. Round 3 used interval periods and obtained the weaker obstruction $f(\alpha)\le \min(\alpha,1-\alpha)$. Round 4 uses a different family of periods and sharpens the admissible upper bound on $f(\alpha)$ to the random-model ceiling $\alpha(1-\alpha)$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 4 does not resolve Erdős Problem #38, but it makes a substantive advance beyond Round 3:

1. it proves, rigorously and not just heuristically, the random-set ceiling \[ f(\alpha)\le \alpha(1-\alpha) \qquad (\alpha\in \mathbb Q\cap(0,1)); \]
2. this strictly improves the Round 3 obstruction \[ f(\alpha)\le \min(\alpha,1-\alpha); \]
3. it introduces a new tool --- discrepancy-minimising rotation of pseudorandom periods --- which is qualitatively different from the earlier interval/block constructions.

In particular, any future positive solution must work with a function $f$ lying below the random-model barrier at every rational density, and any future negative solution will need ideas beyond the one-block periodic obstructions used in the first three rounds.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 58%

9) REFERENCES

• T. F. Bloom, Erdős Problem #38, erdosproblems.com/38, accessed 2026-03-07.
• Formal Conjectures, file FormalConjectures/ErdosProblems/38.lean, checked 2026-03-07.
• C. McDiarmid, On the method of bounded differences, in Surveys in Combinatorics, 1989 (Norwich, 1989), London Math. Soc. Lecture Note Ser. 141, Cambridge Univ. Press, 1989, 148--188.

Erdős Problem #38: a density-exact obstruction for every real density

Independent continuation, 23 September 2026. Source version considered: gpt_pro_5.4/38.tex.

1) Statement and current resolution

For $A\subseteq\mathbb N=\{1,2,\ldots\}$ let \[ d_s(A)=\inf_{n\ge1}\frac{|A\cap[1,n]|}{n}. \] The question asks for a set $B$ which is not an additive basis, but which admits a function $f(\alpha)>0$ for $0<\alpha<1$ such that, for every $A$ of Schnirelmann density $\alpha$ and every $N$, some $b\in B$ satisfies \[ |(A\cup(A+b))\cap[1,N]|\ge(\alpha+f(\alpha))N. \] The current editorial reports a positive solution, with a sparse random $B$ and $f(\alpha)\gg\alpha(1-\alpha)^2$, and labels the problem PROVED (LEAN). Thus the older source's description of the original question as open is outdated. This note does not reconstruct that positive construction. It independently proves an obstruction, extending the supplied rational-density argument to every real density without any regularity assumption on $f$.

2) Audit and objective

The earlier attempt proves the familiar random-set ceiling $f(\alpha)\le\alpha(1-\alpha)$ for rational $\alpha$ using exact-density periodic sets. A periodic set always has rational density, and rational approximation alone would not extend the conclusion: the problem assumes no continuity or monotonicity of $f$.

We close precisely that logical gap. The resulting theorem is an upper bound compatible with the reported positive solution; it is not a disproof of the problem or a claim of a new literature result.

3) A cyclic prefix with controlled correlations

Fix $\alpha\in(0,1)$ and $\varepsilon>0$. For a sufficiently large integer $Q$, put $k=\lceil\alpha Q\rceil$ and $\beta=k/Q\in(0,1)$. There exists $R\subseteq\mathbb Z/Q\mathbb Z$ with $|R|=k$ and \[ |R\cap(R+t)|\ge(\beta^2-\varepsilon)Q\qquad(t\ne0). \tag{1} \] To see this, take independent Bernoulli variables $\xi_x$ of mean $\beta$ and set $X_t=\sum_x\xi_x\xi_{x+t}$. For $t\ne0$, $\mathbb EX_t=\beta^2Q$. Changing one coordinate changes $X_t$ by at most two, including when $2t=0$: in that case there are two equal summands, still giving the bound two. The bounded-differences inequality gives \[ \mathbb P(X_t<\beta^2Q-\varepsilon Q)\le e^{-\varepsilon^2Q/2}. \] For completeness, the needed inequality follows by exposing the independent coordinates in order. The corresponding conditional-expectation martingale difference has conditional range of length at most two. Hoeffding's elementary exponential bound for a centred variable of range length two is $\mathbb E e^{\lambda Z}\le e^{\lambda^2/2}$. Multiplication of these conditional bounds and Markov's inequality give $\mathbb P(X_t-\mathbb EX_t\le-u)\le\inf_{\lambda>0}\exp(-\lambda u+Q\lambda^2/2)=\exp(-u^2/(2Q))$. This argument uses coordinate independence, not independence of the overlapping products.

The binomial variable $Y=\sum_x\xi_x$ has $k$ as a mode, so $\mathbb P(Y=k)\ge1/(Q+1)$. This mode assertion follows directly from the ratio of consecutive binomial probabilities, since $\beta=k/Q$. A union bound gives \[ \mathbb P(\exists t\ne0\text{ violating (1)})\le(Q-1)e^{-\varepsilon^2Q/2}<\frac1{Q+1} \] for large $Q$. Therefore some outcome has both $Y=k$ and $1$. The estimate is uniform in $\beta$, so its dependence on $Q$ creates no issue.

Write the indicator of $R$ as a binary word $w_1,\ldots,w_Q$ and let $D_j=\sum_{i=1}^j(w_i-\beta)$. Rotate the word to start immediately after a minimum of $D_0,\ldots,D_{Q-1}$. Because $D_Q=0$, every prefix of the rotated word has at least $\beta$ times its length ones; this follows by subtracting the minimum discrepancy, including across the cyclic boundary. Rotation preserves $1$. Identify this rotated word with a subset $R'\subseteq[1,Q]$.

4) Exact irrational density from a nonperiodic tail

Define $A\cap[1,Q]=R'$. For $j\ge1$, include $Q+j$ in $A$ precisely when \[ \lceil\alpha j\rceil-\lceil\alpha(j-1)\rceil=1. \] The difference is always zero or one. Thus \[ |A\cap[1,Q+j]|=k+\lceil\alpha j\rceil\ge\alpha(Q+j). \] For prefixes at most $Q$, the rotation gives density at least $\beta\ge\alpha$. All initial densities are therefore at least $\alpha$. On the other hand their limit is $\alpha$, since the fixed initial excess and the ceiling error vanish after division by $Q+j$. Consequently $d_s(A)=\alpha$, whether $\alpha$ is rational or irrational. This exact equality is the essential improvement over a continuity argument.

5) The universal ceiling

At the particular cutoff $N=Q$, no part of the added tail affects $A$ or a positive translate of $A$. If $b\ge Q$, the translate contributes no points. If $1\le b<Q$, deleting wrapped points only decreases the union, so $1$ gives \[ |(A\cup(A+b))\cap[1,Q]|\le |R'\cup(R'+b)|_{\rm cyclic} \le(2\beta-\beta^2+\varepsilon)Q. \] The same upper bound holds for $b\ge Q$, since $|R'|=\beta Q$ and $\beta\le2\beta-\beta^2$. Now $0\le\beta-\alpha<1/Q$ and \[ (2\beta-\beta^2)-(2\alpha-\alpha^2) =(\beta-\alpha)(2-\alpha-\beta)\le2/Q. \] It follows that every positive shift satisfies \[ |(A\cup(A+b))\cap[1,Q]|\le \left(\alpha+\alpha(1-\alpha)+\varepsilon+2/Q\right)Q. \] If a proposed gain at this exact $\alpha$ were greater than $\alpha(1-\alpha)$, choose $\varepsilon$ smaller than half the excess and then choose a sufficiently large $Q$ so that $2/Q$ is smaller than the remaining excess. The displayed bound contradicts the required property even for $B=\mathbb N$. Hence \[ \boxed{f(\alpha)\le\alpha(1-\alpha)\quad\hbox{for every real }0<\alpha<1.} \] No limit of values $f(\beta)$ has been taken: all test sets have the original density $\alpha$ exactly.

6) What this attempt does and does not complete

The ceiling is now proved for the entire stated density domain, with the irrational-density gap removed. The positive existence theorem still requires a sparse set $B$ which works simultaneously for every $A$ and every cutoff. This note does not supply that construction, its probabilistic quantifiers, or a verification that the constructed $B$ fails to be an additive basis.

ATTEMPT UNRESOLVED. This label describes the incomplete reconstruction in this file; the original problem is reported solved affirmatively by the current source. In particular, the ceiling proved here is compatible with a gain of order $\alpha(1-\alpha)^2$ and cannot negate that result.

7) Sources

https://www.erdosproblems.com/38, current statement and attributed resolution, checked 23 September 2026. The older local gpt_pro_5.4/38.tex supplies the rational-density baseline considered above. The extension and all deductions used here are written out; no publication novelty is asserted.

The estimate measures the portion of the original existence problem reconstructed here, rather than correctness of the proved ceiling or the problem's official resolution status.

8) COMPLETION ESTIMATE

COMPLETION: 15%

Erdős Problem #38 --- Round 2 (continuation & gap-closure)

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round 1 isolated one explicit obstruction (a gcd obstruction) and one adversarial family of sets $A$ (periodic block sets) but left two concrete gaps: (i) the gcd obstruction was built on a set $A=d\mathbb N$ whose Schnirelmann density is actually $0$ under the stated convention $\mathbb N=\{1,2,\dots\}$, (ii) the periodic-block calculation was only carried out for "small" shifts $b\le P-L$.

In Round 2 I (a) correct the gcd obstruction with a density-$1/d$ set that does contain $1$, and (b) strengthen the periodic-block calculation to an all-shifts formula (for $\alpha\le 1/2$) on a single period. This yields new necessary structural conditions on any candidate $B$, in particular that $1\in B$ and that $B$ cannot have unbounded multiplicative gaps.

2) ROUND-1 FOUNDATION USED

I rely on the following Round 1 components (without re-proving them):

• Round 1 formal restatement of the problem and notation.
• Round 1 Lemma 38.2 (periodic block sets): for $A_{P,L}=\bigcup_{k\ge 0}\{kP+1,\dots,kP+L\}$ one has $d_s(A_{P,L})=L/P$.
• Round 1 Lemma 38.2 (finite-$N$ upper bound) in the regime $N=KP$ and $1\le b\le P-L$, giving \[ |(A_{P,L}\cup(A_{P,L}+b))\cap[1,N]|\le \left(\frac LP+\frac bP\right)N. \]

I will explicitly flag where Round 1 contained an error and supply a replacement argument.

3) NEW INSIGHT / TOOL (ROUND-2)

• Schnirelmann sensitivity at $1$. Under $\mathbb N=\{1,2,\dots\}$, any $A$ with $1\notin A$ has $d_s(A)=0$. This invalidates the Round 1 use of $A=d\mathbb N$ as a positive-density example.
• All-shifts union formula for block sets when $\alpha\le \tfrac12$. For $A_{P,L}$ with $L\le P/2$ and for every shift $0\le b\le P-1$, one can compute $|(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|$ exactly and deduce the sharp upper bound \[ \frac{|(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|}{P}\le \alpha+\frac{\min(b,P-b)}{P},\qquad \alpha=L/P\le \tfrac12. \]
• New structural consequences for $B$. Specialising to $\alpha=1/2$ forces $1\in B$ (from $N=2$) and forces $B$ to intersect the "middle" interval $[f(1/2)P,\,P-f(1/2)P]$ for every even $P$, implying bounded multiplicative gaps $b_{n+1}/b_n\ll 1/f(1/2)$.

4) ATTACK PLAN (ROUND-2)

1. Fix the Round 1 gcd obstruction. Prove the elementary lemma "$1\notin A\Rightarrow d_s(A)=0$" and replace $A=d\mathbb N$ by $A=\{1\}\cup d\mathbb N$, which has $d_s(A)=1/d$.
2. Close the wrap-around gap for periodic blocks. Compute the union size on a single period $[1,P]$ for all shifts $0\le b<P$ when $L\le P/2$.
3. Extract new necessary conditions on $B$ from $\alpha=1/2$. Apply the problem's property to $A_{P,P/2}$ at $N=P$ to force existence of a shift $b\in B\cap[1,P]$ with $\min(b,P-b)\ge f(1/2)P$. This yields: (i) $1\in B$ (take $P=2$), (ii) bounded multiplicative gaps (take $P$ just below $b_{n+1}$).

These steps strictly strengthen the Round 1 obstruction analysis without assuming anything about additive bases beyond the problem statement.

5) WORK (ROUND-2)

5.1 A basic correction: sets missing $1$ have $d_s=0$

5.2 Corrected gcd obstruction (fixing Round 1 Lemma 38.1)

Round 1 argued with $A=d\mathbb N$, but Lemma lem:missing1 shows $d_s(d\mathbb N)=0$ under the present convention. The obstruction still holds after a minimal repair.

5.3 All-shifts union formula for periodic block sets when $\alpha\le 1/2$

We now strengthen the Round 1 periodic-block computation by removing the restriction $b\le P-L$ when $L\le P/2$ and by working at a single period ($N=P$).

5.4 Immediate forcing of $1\in B$

5.5 Bounded multiplicative gaps forced by $\alpha=1/2$

Set $f_0:=f(1/2)>0$.

5.6 A small generalisation for rational $\alpha\le 1/2$

For later use it is convenient to record the same "middle interval" forcing for any rational density $\alpha\le 1/2$.

6) ADVERSARIAL VERIFICATION

• Check the corrected gcd obstruction. The only delicate point is that $d\mathbb N+b\subseteq d\mathbb N$ (not equality), but this suffices because it implies $A\cup(A+b)=A\cup\{1+b\}$ and hence the union count increases by at most $1$. The computed Schnirelmann density $d_s(\{1\}\cup d\mathbb N)=1/d$ is robust because the ratio is bounded below by $1/d$ for all $N$ and approaches $1/d$ along $N=kd$.
• Quantifiers / "for every $N$" issues. All deductions from $\alpha=1/2$ used the problem requirement at specific even values $N=P$ (not asymptotic $N\to\infty$), so there is no hidden passage to a subsequence.
• Boundary cases in Lemma lem:allshifts. The casework uses $L\le P/2$ to ensure in Case 1 that $P-b\ge L$ and in Case 3 that $b\ge L$; these are exactly where wrap-around could otherwise appear. The endpoints $b=0$ and $b=P-1$ are covered: the formula gives union size $L$ and $L+1$ respectively, matching the direct interval picture.
• Could Proposition Proposition (Middle-interval condition at every even scale) fail because the witness $b$ might be $\ge P$? No: if $b\ge P$ then $(A_{P,P/2}+b)\cap[1,P]=\emptyset$, so the union count is $P/2$, which cannot reach $(1/2+f_0)P$ since $f_0>0$.
• Internal consistency with Round 1. Round 1 Lemma 38.2 remains correct (it crucially uses sets containing $1$), and Lemma lem:allshifts is a strict strengthening on the special case $N=P$ and $\alpha\le 1/2$. The only Round 1 piece that needed correction was the use of $A=d\mathbb N$ as a positive-$d_s$ example.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED). We have not resolved whether such a non-basis $B$ exists. However, Round 2 makes two substantive advances beyond Round 1:

1. The gcd obstruction is repaired in a way compatible with the stated Schnirelmann density convention (Lemma Lemma (Corrected gcd obstruction)).
2. New necessary structural conditions are proved for any candidate $B$:
   • $1\in B$ (Proposition Proposition);
   • $B$ must hit middle intervals at every even scale (Proposition Proposition (Middle-interval condition at every even scale)), implying bounded multiplicative gaps (Corollary Corollary (No unbounded multiplicative gaps)).

These conditions rule out broad classes of natural non-bases (e.g. lacunary sequences) and tighten the search space for a potential counterexample.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%.

9) REFERENCES

No external references were used beyond the problem statement and Round 1 results.