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Source. U. V. Linnik, “On Erdös's theorem on the addition of numerical sequences,” English Sections 2–5, printed pp. 70–77 (PDF pp. 4–11). The preliminary lemmas and the external Vinogradov and large-sieve inputs are on [[integer_sequences/linnik_1942_erdos_theorem_addition_numerical_sequences/lemmas|the preliminary-lemmas page]]. The exact scan is identified in [[integer_sequences/linnik_1942_erdos_theorem_addition_numerical_sequences/_index|the source index]].

The printed result

The paper's main result is unnumbered. Its definition (p. 67): an essential component is a sequence $\Phi$ whose sum with every sequence $F$ of positive density at least $\beta$ has density at least $\beta+\varphi(\beta)$, where $\varphi(\beta)$ depends only on $\beta$ and not on other properties of $F$, for $\beta<1$. A basic sequence is one for which there is an integer $a_0$ such that every natural number is a sum of at most $a_0$ of its terms. The density is Schnirelmann's, as the fourth lemma's hypothesis $\Psi(N)\geq\beta N$ shows. After constructing $\Phi_1$ and $\Phi_2$ (p. 70), the paper states: "We form now the sequence $2\Phi_1+\Phi_2=\Phi_1+\Phi_1+\Phi_2$ by ordinary rules and so obtain a sequence $\Phi$. The sequence $\Phi$ will be a non-basic essential component." (p. 70). Non-basishood is proved in § 2 (pp. 70–71) and the essential-component property in §§ 3–5 (pp. 71–77).

"By ordinary rules" is read here as the addition of sequences in Schnirelmann's theory, where a sum contains its summands. On that reading $1\in\Phi$ and $F\subseteq F+\Phi$, which p. 77 uses. This page gives a reconstruction of the same power-set and Fourier argument, with the changes specified below. It writes all sums as Minkowski sums and adjoins $\{0,1\}$ explicitly, so the proved component is $\widehat\Phi$ below.

Statement and addition convention

For $F\subseteq\mathbb Z_{>0}$, write

$$\sigma(F)=\inf_{N\in\mathbb Z_{\geq1}} \frac{|F\cap[1,N]|}{N}.$$

There is a fixed set $\widehat\Phi\subseteq\mathbb Z_{\geq0}$, defined below, which is not an additive basis of any finite order and has the following property. For every $0<\beta<1$ there is $\varphi(\beta)>0$ such that, for every $F\subseteq\mathbb Z_{>0}$ with $\sigma(F)\geq\beta$ and every integer $N\geq1$,

$$|(F+\widehat\Phi)\cap[1,N]| \geq(\beta+\varphi(\beta))N. \tag{E}$$

All sums here are ordinary Minkowski sums. Equivalently, one can use the positive sequence $\widehat\Phi\setminus\{0\}$ and explicitly retain $F$ when adding it. The conclusion is a Schnirelmann-density bound at every cutoff, not just a lower asymptotic-density bound.

This is the essential-component assertion studied by Linnik. It concerns the union of all translates by the component; it does not by itself give the stronger single-shift assertion in Problem 38.

The power sets and the finite augmentation

Use the English construction on p. 70:

$$c_0=\exp(14^{20}),\qquad N_0=\left\lfloor \exp\bigl((\log c_0)^{10/9}\bigr) \right\rfloor,\qquad N_j=N_0^{\,2^j}\quad(j\geq0).$$

Put

$$n_j=\lfloor(\log N_j)^{1/10}\rfloor,\qquad P_j=N_j^{1/n_j},\qquad d_j=\lfloor n_j/2\rfloor.$$

The starting number is more than sufficient to ensure $n_j\geq3$. Define sets of power values, rather than bounds on their bases:

$$\begin{aligned} A_0&=\{x^{n_0}:x\in\mathbb Z_{\geq1},\ 1\leq x^{n_0}\leq N_0\},\\ A_1&=\{x^{n_1}:x\in\mathbb Z_{\geq1},\ 1\leq x^{n_1}\leq N_1\},\\ A_j&=\{x^{n_j}:x\in\mathbb Z_{\geq1},\ N_{j-2}\leq x^{n_j}\leq N_j\}\quad(j\geq2). \end{aligned}$$

Let $B_0,B_1,B_j$ have the same respective value cutoffs, with degree $d_j$ in place of $n_j$. In particular, the half-degree sets include both $B_0$ and $B_1$.

Set

$$\Phi_1=\bigcup_{j\geq0}(A_0+A_1+\cdots+A_j)\ \cup\ \bigcup_{l\geq0}A_l, \qquad \Phi_2=\bigcup_{j\geq0}B_j,$$

where the $j=0$ summand in the first union is $A_0$. This spells out the English definition, which takes the prefix sums and also each set $A_l$ by itself; p. 70 prints this subscript as an italic $l$, not the digit $1$. The restriction of $\Phi_1$ to $[1,N_k]$ receives no contribution from prefixes ending at $j\geq k+2$: such a prefix contains a term from $A_{k+2}$ at least $N_k$ and additional positive terms. Thus it agrees with the cutoff prescription on p. 70. The printed fifth lemma, where $N_{k-1}\leq N<N_k$, places $\lfloor(\beta N/40)^{1/n_k}\rfloor^{n_k}$ in $\Phi_1$ (p. 73); for large $N$ this power lies in the single set $A_k$.

Finally define

$$\Phi=\Phi_1+\Phi_1+\Phi_2, \qquad \widehat\Phi=\{0,1\}\cup\Phi. \tag{C}$$

The positive sets satisfy $1\in\Phi_1\cap\Phi_2$, and the Minkowski sum $\Phi$ has $\min\Phi=3$, so it does not contain $1$, which the paper's convention supplies (p. 77). The finite adjunction in (C) explicitly supplies $F\subseteq F+\widehat\Phi$ and $F+1\subseteq F+\widehat\Phi$, as needed in the density argument. It does not change any of the power sets.

Non-basishood, including the degree floors

Let $k\geq1$ and put

$$R_k=\prod_{i=0}^{k}(1+P_i).$$

The total number of choices for prefixes ending at $0,\ldots,k$ is at most $R_k$: their products of cardinalities occur among the nonnegative terms in its expansion. The single sets $A_0,\ldots,A_k$ contribute at most $R_k$ more points, since $|A_i|\leq P_i$. A prefix ending at $k+1$ can use at most $N_k^{1/n_{k+1}}\leq P_k$ values from its last set, since the degrees are nondecreasing; by the same bound the single set $A_{k+1}$ contributes at most $P_k$ points. The single set $A_{k+2}$ can contribute only the value $N_k$, and later sets contribute nothing. Since $P_k+1\leq R_k$,

$$|\Phi_1\cap[1,N_k]|\leq(P_k+3)R_k. \tag{C1}$$

For every integer $n\geq3$, $\lfloor n/2\rfloor\geq n/3$. Thus

$$|B_i|\leq N_i^{1/d_i}\leq P_i^3.$$

At cutoff $N_k$, the set $B_{k+1}$ contributes at most $N_k^{1/d_k}\leq P_k^3$ points. The set $B_{k+2}$ can contribute only the value $N_k$, and later sets contribute nothing. Therefore

$$|\Phi_2\cap[1,N_k]| \leq 1+P_k^3+\sum_{i=0}^{k}P_i^3=:D_k. \tag{C2}$$

Since all summands in $\Phi$ are positive, (C1) and (C2) give

$$|\widehat\Phi\cap[0,N_k]| \leq2+\bigl((P_k+3)R_k\bigr)^2D_k. \tag{C3}$$

To estimate this quantity, write $L_i=\log N_i=2^iL_0$. The floor in $n_i$ gives $n_i\geq L_i^{1/10}/2$, so

$$\log P_i\leq2L_i^{9/10},\qquad \sum_{i=0}^{k}L_i^{9/10} \leq\frac{L_k^{9/10}}{1-2^{-9/10}}.$$

It follows directly from (C1)–(C3) that

$$\log|\widehat\Phi\cap[0,N_k]| =O(L_k^{9/10}+k+\log(k+3)) =O(L_k^{9/10}). \tag{C4}$$

In particular, $|\widehat\Phi\cap[0,N_k]|=N_k^{o(1)}$. For any fixed positive integer $h$, every element of $h\widehat\Phi$ below $N_k$ uses only elements of $\widehat\Phi\cap[0,N_k]$. There are at most $|\widehat\Phi\cap[0,N_k]|^h=o(N_k)$ ordered choices. Thus $h\widehat\Phi$ cannot contain all sufficiently large integers. This also proves non-basishood of the unaugmented $\Phi$.

The powers $P_i^3$ above account for odd $n_i$; the printed $2\sum P_i^2$ bound does not. The geometric series has sum $(1-2^{-9/10})^{-1}>2$, so the printed intermediate constant $32$ on p. 71 also needs replacement. Neither numerical assertion is used here.

Parameters and the power-sum estimates

Fix $0<\beta<1$ for the rest of the proof. Put

$$\beta_1=\frac{1-\beta}{2},\qquad D_\beta=\frac1{\beta_1}+\frac{200}{\beta\beta_1}, \qquad K=10^6D_\beta.$$

We first record precisely which power sums permit the sufficient estimate (W) on the preliminary-lemmas page. Let

$$p_j=\lfloor P_j\rfloor,\qquad q_j=\left\lceil N_{j-2}^{1/n_j}\right\rceil-1\quad(j\geq2).$$

The bases defining $A_j$ are then $q_j+1,\ldots,p_j$. As $j\to\infty$,

$$n_j\sim L_j^{1/10},\qquad \log p_j\sim L_j^{9/10},\qquad \frac{q_j}{p_j}\longrightarrow0. \tag{P1}$$

The last limit follows by comparing the logarithms of the lower and upper roots, $L_j/(4n_j)$ and $L_j/n_j$; integer rounding has a vanishing relative effect. Hence eventually

$$1\leq q_j\leq p_j/2,\qquad 14\leq n_j\leq2(\log p_j)^{1/9}. \tag{P2}$$

Also, for all sufficiently large $j$,

$$p_j<p_{j+1}\leq p_j^{\,n_j-1}. \tag{P3}$$

Indeed, $\log p_{j+1}/\log p_j\to2^{9/10}>1$, whereas $(n_j-1)\log p_j\sim L_j$ and $\log p_{j+1}=O(L_j^{9/10})$.

There is a uniform version for the last, truncated sum. If $N_{r-1}\leq H<N_r$ and

$$p=\lfloor H^{1/n_r}\rfloor,\qquad q=\left\lceil N_{r-2}^{1/n_r}\right\rceil-1,$$

then, uniformly in that range of $H$, as $r\to\infty$,

$$1\leq q\leq p/2,\qquad 14\leq n_r\leq2(\log p)^{1/9}. \tag{P4}$$

For the first assertion, the logarithmic separation of the two roots is at least $L_r/(4n_r)\to\infty$. For the second, $\log H\geq L_r/2$ gives $n_r/(\log p)^{1/9}\leq2^{1/9}+o(1)<2$. These estimates also show that $p\to\infty$ uniformly.

Choose $j_0\geq2$ sufficiently large that (P2)–(P3) hold for all $j\geq j_0$, $p_{j_0}\geq P_*$, and

$$\exp\bigl(-\sqrt{\log p_{j_0}}\bigr)\leq\frac1K.$$

Here $P_*$ is the absolute threshold in (W). Set

$$b_0=p_{j_0},\qquad b=\lceil10Kb_0\rceil,\qquad \delta=\frac{\beta\beta_1}{1600b}. \tag{P5}$$

These constants depend only on $\beta$; $b$ is an integer greater than one. No predecessor index is needed at the initial crossing.

A large cutoff with too little growth

Let $\sigma(F)\geq\beta$ and put $F_1=F+\widehat\Phi$. We will show that for all sufficiently large integers $N$, with a threshold depending only on $\beta$,

$$|F_1\cap[1,N]|\geq(\beta+\delta)N. \tag{G}$$

Suppose to the contrary that the reverse strict inequality holds. Since $\delta<\beta_1$, the complement of $F_1$ below $N$ has more than $\beta_1N$ points. Let

$$\mathcal M_N= \{m\in[1,N]\cap\mathbb Z:m\notin F_1,\ m\geq\beta_1N/2\}, \qquad Z_1=|\mathcal M_N|.$$

Fewer than $\beta_1N/2$ positive integers lie below $\beta_1N/2$. Therefore

$$Z_1>\frac{\beta_1N}{2}. \tag{G1}$$

Take the smaller cutoff

$$H=\left\lfloor\frac{\beta_1N}{100}\right\rfloor. \tag{G2}$$

For sufficiently large $N$, $H\geq\beta_1N/200$ and $H>4b/\beta$. Apply the corrected fourth lemma at $H$, with $C=b$ and $\varepsilon=\beta/2$. Its first alternative would give

$$|(F\cup(F+1))\cap[1,H]|-|F\cap[1,H]| \geq\frac{\beta H}{4b} \geq\frac{\beta\beta_1N}{800b}=2\delta N.$$

All these new points belong to $F_1$, and $F\subseteq F_1$. This would contradict $|F_1\cap[1,N]|<(\beta+\delta)N$.

Consequently the set

$$\mathcal G_N= \{f\in F\cap[1,H]:\{f,f+1,\ldots,f+b\}\subseteq F\cap[1,H]\}$$

has cardinality

$$Z_2=|\mathcal G_N|>\frac{\beta H}{2} \geq\frac{\beta\beta_1N}{400}. \tag{G3}$$

This directly counts good starts at the cutoff where they are used; it does not discard an uncontrolled second half of an interval.

Choose the unique $k$ with $N_{k-1}\leq N<N_k$, and put

$$X_1=\left\lfloor N^{1-11/(10n_k)}\right\rfloor. \tag{G4}$$

For sufficiently large $N$,

$$N^{3/4}<X_1<\frac{N}{(\log N)^2}.$$

Here $n_k\to\infty$, while $(\log N)/n_k$ grows on the order of $(\log N)^{9/10}$, faster than $\log\log N$. These facts justify both inequalities, including the floor. Apply the difference form of the third lemma to $\mathcal M_N,\mathcal G_N\subseteq[1,N]$, with $\gamma_0=\beta\beta_1/800$. We obtain a prime $p\in[X_1/2,X_1]$ such that every residue $v$ satisfies

$$\#\{(m,f)\in\mathcal M_N\times\mathcal G_N: m-f\equiv v\pmod p\} \geq c_{\mathrm{rep}}\frac{Z_1Z_2}{p}, \qquad c_{\mathrm{rep}}=\frac{0.996}{16}. \tag{G5}$$

Fifth lemma: an interval in the difference set

Define

$$\mathcal D_N= \bigl(\mathcal M_N-(F\cap[1,N]) -(\Phi_1\cap[1,N])-(\Phi_1\cap[1,N])\bigr)\cap[1,N].$$

For all sufficiently large $N$ under the preceding supposition, $\mathcal D_N$ contains all integers in an interval $[Y_1,Y_2]$ with

$$1\leq Y_1<Y_2\leq N,\qquad Y_2-Y_1\geq X_1/4. \tag{L}$$

This is the form of Linnik's fifth lemma required below. We prove it with explicit supports and cardinalities.

Omitted values and the comparison sum

Suppose (L) fails, and let $s=\lceil X_1/4\rceil$ and $w=\lfloor N/p\rfloor$. Every interval of $s+1$ consecutive integers inside $[1,N]$ then has a value omitted by $\mathcal D_N$. For $j=1,\ldots,w$, choose one such omitted value

$$w_j\in[jp-s,jp]\setminus\mathcal D_N.$$

When $X_1\geq8$, these intervals lie in $[1,N]$ and are disjoint: $s+1\leq p$. In particular, the $w_j$ are distinct and

$$0\leq jp-w_j\leq s. \tag{L1}$$

Let $r$ be determined by $N_{r-1}\leq H<N_r$. For $N$ sufficiently large we have $r>j_0$. Take

$$\begin{aligned} S_1(\alpha)&=\sum_{m\in\mathcal M_N}e(\alpha m),& S_2(\alpha)&=\sum_{f\in\mathcal G_N}e(-\alpha f),\\ U(\alpha)&=\sum_{u=0}^{b}e(-\alpha u),& T_j(\alpha)&=\sum_{t\in A_j}e(-\alpha t)\quad(0\leq j<r),\\ T_r(\alpha)&=\sum_{\substack{x\in\mathbb Z_{\geq1}\\ N_{r-2}\leq x^{n_r}\leq H}}e(-\alpha x^{n_r}),& W(\alpha)&=\sum_{j=1}^{w}e(-\alpha w_j),\\ Q(\alpha)&=\sum_{x=0}^{w}e(-\alpha px).&& \end{aligned}$$

Write

$$R(\alpha)=U(\alpha)\prod_{j=0}^{r}T_j(\alpha), \qquad A=R(0)=(b+1)\prod_{j=0}^{r}T_j(0).$$

All the power sums are nonempty for these large cutoffs. A selection of their terms has sum

$$\phi=t_0+\cdots+t_r\in A_0+\cdots+A_r\subseteq\Phi_1.$$

Since $N_{i+1}=N_i^2$ and $N_0\geq2$,

$$\phi\leq\sum_{j=0}^{r-1}N_j+H \leq2N_{r-1}+H\leq3H. \tag{L2}$$

This is the truncation needed in the counting argument.

Use the fixed value $M=1\in\Phi_1$ and consider

$$I=\int_0^1 S_1S_2RW\,e(-\alpha)\,d\alpha, \qquad J=\int_0^1 S_1S_2RQ\,e(-\alpha)\,d\alpha. \tag{L3}$$

Orthogonality shows that $I$ counts the ordered choices satisfying

$$w_j=m-(f+u)-\phi-1.$$

Here $f+u\in F\cap[1,H]$, $\phi\in\Phi_1\cap[1,N]$, and $1\in\Phi_1\cap[1,N]$. Thus any counted $w_j$ would belong to $\mathcal D_N$, contrary to its selection. Hence $I=0$.

For $J$, fix $u,t_0,\ldots,t_r$. Every pair $(m,f)$ satisfying the appropriate congruence from (G5) gives

$$m-(f+u)-\phi-1=px. \tag{L4}$$

Indeed, the left side is an integer between $0$ and $N$: its lower bound is

$$\frac{\beta_1N}{2}-4H-1>0$$

for sufficiently large $N$, by (G2) and (L2), while its upper bound is at most $m\leq N$. Therefore its quotient by $p$ is one of $0,\ldots,w$. This proves the lower bound

$$J\geq c_{\mathrm{rep}}\frac{Z_1Z_2}{p}A. \tag{L5}$$

Denominator coverage and the minor arcs

Put

$$P=\lfloor H^{1/n_r}\rfloor,\qquad \tau=P^{n_r-1}.$$

Both are integers. In view of (P4), the sufficient estimate (W) applies to $T_r$ for denominators in $[P,\tau]$. It applies to $T_j$, $j_0\leq j<r$, for denominators in $[p_j,p_j^{n_j-1}]$. Take $N$ large enough that $P\geq b_0$ and that all the truncated conditions (P4) hold. Each applicable estimate then bounds its factor by at most its cardinality divided by $K$.

By (P3), the intervals $[p_j,p_j^{n_j-1}]$ overlap successively. Moreover,

$$P\leq p_r\leq p_{r-1}^{n_{r-1}-1}.$$

The last interval $[P,\tau]$ therefore overlaps that chain. Every denominator in $[b_0,\tau]$ is covered by at least one of these power sums. The initial factors with $j<j_0$ require only their trivial cardinality bounds.

Dirichlet's approximation theorem, with the integer cutoff $\tau$, gives, for every $\alpha\in\mathbb R/\mathbb Z$, coprime integers $a,q$ with

$$1\leq q\leq\tau,\qquad \left|\alpha-\frac aq\right|\leq\frac1{q\tau}. \tag{L6}$$

For the minor arcs $\|\alpha\|>2/\tau$, where $\|\alpha\|$ denotes distance to the nearest integer, this cannot have $q=1$. If $q\geq b_0$, the preceding denominator coverage gives cancellation in one power sum; (L6) also implies the required error at most $1/q^2$.

If $2\leq q<b_0$, then for $\tau\geq2$,

$$\|\alpha\|\geq\frac1q-\frac1{q\tau}\geq\frac1{2q}.$$

The geometric-sum estimate gives

$$|U(\alpha)|\leq\frac1{2\|\alpha\|}\leq q<b_0 \leq\frac{b+1}{10K}.$$

Thus in all cases

$$|R(\alpha)|\leq A/K \qquad(\|\alpha\|>2/\tau). \tag{L7}$$

Parseval and the arithmetic-geometric mean inequality give

$$\begin{aligned} \int_0^1|S_1S_2|\,d\alpha &\leq\frac{Z_1+Z_2}{2}\\ &=\frac{Z_1Z_2}{2}\left(\frac1{Z_1}+\frac1{Z_2}\right) \leq D_\beta\frac{Z_1Z_2}{N}, \end{aligned} \tag{L8}$$

using (G1) and (G3). Since $|W|\leq w$ and $|Q|\leq w+1\leq2w$, the absolute values of the minor-arc parts of $I$ and $J$ are at most, respectively,

$$\frac{D_\beta}{K}\frac{AZ_1Z_2w}{N}, \qquad \frac{2D_\beta}{K}\frac{AZ_1Z_2w}{N}. \tag{L9}$$

The major arcs and the distinct scales

The needed major-arc approximation follows from $X_1/\tau\to0$. Here the definitions of $X_1$ and $\tau$ use different cutoffs and possibly different degree indices; they are not equated.

For fixed $\beta$, (G2) implies $H=(\beta_1/100)N+O(1)$. Eventually $H\geq N_{k-2}$, because $N\geq N_{k-1}=N_{k-2}^2$. Since $H<N<N_k$, this gives $r\in\{k-1,k\}$. The degree floors satisfy

$$\frac{n_k}{n_r}\leq1.08$$

for all sufficiently large $k$, because the only nontrivial limiting ratio is $2^{1/10}<1.08$. Integer root rounding gives

$$\log\tau=(1-1/n_r)\log H+o(1).$$

Combining this with (G4),

$$\begin{aligned} \log(X_1/\tau) &=\left(\frac1{n_r}-\frac{1.1}{n_k}\right)\log N -(1-1/n_r)\log(\beta_1/100)+o(1)\\ &\leq-\frac{\log N}{50n_k}+O_\beta(1) \longrightarrow-\infty. \end{aligned} \tag{L10}$$

Also $w\geq\lfloor N/X_1\rfloor\to\infty$. On $\|\alpha\|\leq2/\tau$, pair the terms of $W$ with the nonconstant terms of $Q$. By (L1),

$$|Q(\alpha)-W(\alpha)| \leq1+\frac{4\pi ws}{\tau}\leq\frac wK \tag{L11}$$

for all sufficiently large $N$. Indeed, $1/w+4\pi s/\tau\to0$ uniformly for primes $p\in[X_1/2,X_1]$.

Using the trivial bound $|R|\leq A$ and (L8), the major-arc contribution to $J-I$ is at most

$$\frac{D_\beta}{K}\frac{AZ_1Z_2w}{N}.$$

Together with (L9), this gives

$$|J-I|\leq\frac{4D_\beta}{K}\frac{AZ_1Z_2w}{N}. \tag{L12}$$

But $1/p\geq w/N$, so (L5) and (L12) imply

$$I\geq \left(c_{\mathrm{rep}}-\frac{4D_\beta}{K}\right) \frac{AZ_1Z_2w}{N}>0.$$

Here $c_{\mathrm{rep}}=0.996/16$ and $4D_\beta/K=4\cdot10^{-6}$. This contradicts $I=0$, proving (L).

Half-degree powers meet the interval

We justify the power-gap assertion including transitions between blocks. For a real $z\in[N_{j-1},N_j)$ with $j\geq2$, put

$$b_z=\lfloor z^{1/d_j}\rfloor^{d_j}.$$

For all sufficiently large $j$, uniformly in this range of $z$, $b_z\geq z/2\geq N_{j-2}$. To see the first inequality, write $x=z^{1/d_j}$ and use

$$\frac{b_z}{z}\geq(1-1/x)^{d_j}\geq1-\frac{d_j}{x}\geq\frac12.$$

The last inequality holds uniformly since $x\geq N_{j-1}^{1/d_j}$ grows faster than $d_j$. The second follows from $N_{j-1}=N_{j-2}^2$ and $N_{j-2}\geq2$. Thus $b_z\in B_j\subseteq\Phi_2$. The mean value theorem also gives

$$0\leq z-b_z\leq d_jz^{1-1/d_j}. \tag{H1}$$

When $1\leq z\leq N$ and $N_{k-1}\leq N<N_k$, the relevant index $j$ is at most $k$. The degrees $d_j$ are nondecreasing. Hence the right side of (H1) is at most

$$\Delta_N=d_kN^{1-1/d_k}.$$

The finitely many initial ranges not covered by the uniform argument are contained in some fixed $[1,N_*]$. They can use $1\in B_0$ as the preceding element; for large $N$, $\Delta_N\geq N_*$. We have therefore proved that, for every $z\in[1,N]$, some $b_z\in\Phi_2$ satisfies $z-\Delta_N\leq b_z\leq z$. This argument chooses a power in a valid overlapping block at each cutoff; it does not assume that adjacent blocks have matching endpoints.

Since $d_k=\lfloor n_k/2\rfloor\leq n_k/2$, (G4) yields

$$\frac{\Delta_N}{X_1} \leq(1+o(1))d_k \exp\left(-\frac{0.9\log N}{n_k}\right) \longrightarrow0. \tag{H2}$$

For large $N$, apply this preceding-power assertion to the upper endpoint $Y_2$ in (L). As $\Delta_N<X_1/4\leq Y_2-Y_1$, it supplies $\phi_2\in\Phi_2\cap[Y_1,Y_2]$. By the definition of $\mathcal D_N$,

$$\phi_2=m-f-\phi_1-\phi'_1$$

for some $m\in\mathcal M_N$, $f\in F$, and $\phi_1,\phi'_1\in\Phi_1$. Consequently

$$m=f+\phi_1+\phi'_1+\phi_2\in F+\Phi\subseteq F_1,$$

contrary to $m\in\mathcal M_N$. This proves (G).

Every cutoff and the density increment

All thresholds above depend only on $\beta$: the fourth lemma uses $b,\beta$, the prime-selection lemma uses $\gamma_0=\beta\beta_1/800$, and the remaining conditions follow from the displayed uniform limits and the fixed power construction. Choose an integer $N_\beta\geq1$ beyond all of them. Then (G) holds for every $N>N_\beta$ and every $F$ with $\sigma(F)\geq\beta$.

For $1\leq N\leq N_\beta$, positive Schnirelmann density implies $1\in F$. If $F$ omits a point of $[1,N]$, its first omitted point has a predecessor in $F$, so it belongs to $F+1$. Since $\{0,1\}\subseteq\widehat\Phi$,

$$|F_1\cap[1,N]|\geq|F\cap[1,N]|+1\geq\beta N+1.$$

If no point is omitted, the count is $N$. Set

$$\varphi(\beta)= \min\left\{\delta,\frac1{N_\beta},1-\beta\right\}>0.$$

The two small-cutoff cases and (G) prove (E) for every positive integer $N$. Together with (C4), this proves the asserted non-basic essential-component construction. $\square$

Relation to the printed proof

The reconstruction retains the English power sets, the prime-selection lemma, the omitted-value comparison, Weyl cancellation on overlapping denominator ranges, and the half-degree covering argument. The following changes are substantive bookkeeping corrections, not identities asserted by the scan.

• Pp. 67, 70 and 77: addition. The paper adds sequences "by ordinary rules", read here as Schnirelmann's addition, which keeps the summands and so gives $1\in\Phi$ and $F\subseteq F+\Phi$ (p. 77). Minkowski sums of the positive sets give $\min\Phi=3$, so the stated theorem adjoins $\{0,1\}$ explicitly as $\widehat\Phi$.
• P. 70: construction. The cutoffs bound the powers, and all the half-degree blocks start at $B_0$. The English prefix union is retained exactly. Bounds on the bases and omission of $B_0,B_1$ were errors in the earlier compilation, not in this source definition.
• Pp. 68–70: preliminary estimates. The normalized Weyl estimate used here is the proved sufficient specialization (W); the full printed first-lemma parameter range is not certified here. The fourth lemma needs its terminal boundary and a threshold depending on $\varepsilon$, as proved on the preliminary-lemmas page.
• P. 71: counts and initial index. Odd degrees require a count such as $P_i^3$, and the geometric series cannot be bounded by two. The second printed crossing inequality uses $P_{k_0-1}$, not $\sqrt{\log P_{k_0}}-1$. Choosing a sufficiently large $j_0\geq2$ avoids an undefined predecessor and supplies all needed estimates.
• P. 72: retained mass. The printed second truncation of good starts does not justify its asserted $Z_2$ bound. Here the corrected fourth lemma is applied directly at $H$ and yields (G3). Also $Z_1$ counts the restricted set $\mathcal M_N$, not the whole complement.
• Pp. 72–76: the two scales. The source uses $X_1=N^{1-1.1/n_k}$ and later $\tau=(\beta_1N/10)^{1-1/n_k}$. Its fifth-lemma display on p. 72 has an inconsistent factor $1/2$ versus $1/4$; Section 5 uses the latter. Here integer $X_1$ retains the source's $N$ scale, while $\tau$ belongs to the explicitly truncated $H$ scale. Equation (L10) proves the comparison actually needed.
• Pp. 73–74: signs, membership and positivity. The printed $S_2$ has the wrong sign for the equation it is said to count; the negative sign is used here. As printed, the omitted values already lie outside the difference set: the overlined $\in$ on p. 73 denotes non-membership. The product includes $T_0$, so its sums lie in the stated prefix union. The fixed value $M=1\in\Phi_1$ replaces the printed power at the $\beta N/40$ scale. Full lower blocks and a final block truncated at $H$ satisfy (L2), which proves (L4). The source's displayed supports up to $N$ do not imply its claimed positivity for every choice.
• Pp. 76–77: half-degree gaps. The printed exponent on p. 77 is $1-1/n'_k$, with $n'_k=\lfloor n_k/2\rfloor$. Dropping that prime was a compilation error. Equations (H1)–(H2) supply the floor and block-transition details omitted in the short source argument.
• P. 78: Russian summary. It prints the smaller starting value $\lfloor\exp(14^{20})\rfloor$ and a compressed description of $\Phi_1$ differing from the English prefix union. Those formulas are not mixed into the English proof.

Bears on. Problem 38: the result gives a set that is not a basis of any finite order whose sum with every set of Schnirelmann density at least $\beta\in(0,1)$ gains a fixed $\varphi(\beta)>0$ in density at every cutoff. The sum uses all elements of the set at once, whereas Problem 38 asks that for every $A$ and every cutoff $N$ a single element $b$ give $A\cup(A+b)$ the gain up to $N$, so the result does not by itself answer the problem.