Erdős Problem #5

ROUND-6 OBJECTIVE

Path chosen: (C) obstruction / correction via a new structural reformulation and a stronger strategy-elimination theorem.

Round--5 already showed that one-hole model sets can satisfy all currently known coarse structural properties of the normalized-prime-gap limit-point set $S$ while still failing to be dense. The remaining loophole is that perhaps density might still follow once one excludes single-hole pathologies and insists on a more complicated complement --- for instance infinitely many holes, infinite complement measure, or more realistic global geometry.

The aim of Round--6 is to close that loophole rigorously. We do this in two steps.

1. We prove a new complement-side reformulation of the $k$-point difference property $(P_k)$. For $k=4$, this says that $A\subseteq[0,\infty)$ has $(P_4)$ if and only if its complement contains no six-term additive configuration \[ x,\ y,\ z,\ x+y,\ y+z,\ x+y+z. \]
2. Using that reformulation, we construct lacunary multi-hole countermodels: closed, non-dense sets with infinitely many disjoint holes, asymptotic density $1$, arbitrarily small exact syndetic constant, and arbitrarily long initial segment containment, while still satisfying $(P_4)$.

This strictly advances beyond Round--5: the obstruction is no longer confined to one-gap or finite-measure-complement models.

ROUND-5 FOUNDATION USED

We use the following vetted Round--5 (and earlier) results.

• (Round--1) The actual prime-gap limit-point set $S$ is closed.
• (Round--2) The limit-point set for $(p_{n+1}-p_n)/\log n$ equals the classical one for $(p_{n+1}-p_n)/\log p_n$, so results of Banks--Freiberg--Maynard, Pintz, and Merikoski apply directly to the present problem.
• (Round--3) Relative-gap lemma: if a set $A\subset[0,\infty)$ has $(P_k)$ and $(a,b)\cap A=\emptyset$ with $a>0$, then $b\le (k-1)a$.
• (Round--5) Exact one-gap characterization: for \[ A_{a,b}=[0,a]\cup[b,\infty), \] property $(P_k)$ is equivalent to $b\le (k-1)a$, and for $k=4$ this is equivalent to the sharp initial-interval measure bound \[ \mu(A_{a,b}\cap[0,T])\ge T/3\qquad (T>0). \]
• (Imported earlier for the genuine set $S$) Merikoski proves $(P_4)$, the lower bound $\mu(S\cap[0,T])\ge T/3$ for all $T\ge 0$, and bounded gaps; Pintz gives $[0,c]\subseteq S$ for some ineffective $c>0$.

NEW INSIGHT / TOOL (ROUND-6)

The genuinely new ingredient is a complement reformulation of $(P_k)$. Let $A\subseteq[0,\infty)$ and let \[ B:=[0,\infty)\setminus A. \] Then $A$ has $(P_k)$ if and only if there do not exist nonnegative reals \[ x_1,\dots,x_{k-1} \] such that every contiguous sum \[ x_i+x_{i+1}+\cdots+x_j\qquad (1\le i\le j\le k-1) \] lies in $B$.

For $k=4$, this becomes the explicit additive-pattern exclusion: \[ A\text{ has }(P_4) \iff \text{there do not exist }x,y,z\ge 0\text{ with }x,y,z,x+y,y+z,x+y+z\in B. \]

This makes it possible to construct complements $B$ with infinitely many short holes placed so far apart that the total sum of any three hole-points necessarily lands between two holes, and hence outside $B$. That yields a new infinite-hole family of countermodels.

ATTACK PLAN (ROUND-6)

After Round--5, the exact unresolved logical gap was this:

Could density perhaps still be forced from the known coarse properties if one rules out the one-hole geometry used in earlier countermodels?

Round--6 addresses this by proving:

1. a structural lemma translating $(P_k)$ into a forbidden additive pattern in the complement;
2. a general lacunary-hole construction with infinitely many gaps satisfying $(P_k)$;
3. a specialization to $k=4$ showing that even closedness, $(P_4)$, arbitrarily long initial segment containment, arbitrarily small exact syndetic constant, asymptotic density $1$, and infinitely many complementary components still do not imply density.

Thus any proof of Erdős' conjecture must use arithmetic structure beyond these coarse invariants.

WORK (ROUND-6)

Definition

For an integer $k\ge 2$, a set $A\subseteq[0,\infty)$ has property $(P_k)$ if for every real numbers \[ \beta_1\le \beta_2\le \cdots\le \beta_k \] we have \[ A\cap\{\beta_j-\beta_i:1\le i<j\le k\}\neq\emptyset. \]

Lemma 1 (complement reformulation of $(P_k)$)

Let $A\subseteq[0,\infty)$ and let $B=[0,\infty)\setminus A$. Then $A$ has $(P_k)$ if and only if there do not exist nonnegative reals $x_1,\dots,x_{k-1}$ such that \[ \sum_{t=i}^{j}x_t\in B \qquad\text{for every }1\le i\le j\le k-1. \]

Proof. Suppose first that $A$ fails $(P_k)$. Then there exist reals \[ \beta_1\le \beta_2\le \cdots\le \beta_k \] for which every pairwise difference $\beta_j-\beta_i$ belongs to $B$. Set \[ x_r:=\beta_{r+1}-\beta_r\ge 0\qquad (1\le r\le k-1). \] Then for every $1\le i\le j\le k-1$ we have \[ \sum_{t=i}^{j}x_t = (\beta_{i+1}-\beta_i)+(\beta_{i+2}-\beta_{i+1})+\cdots+(\beta_{j+1}-\beta_j) =\beta_{j+1}-\beta_i\in B. \] Thus all contiguous sums lie in $B$.

Conversely, suppose there exist nonnegative reals $x_1,\dots,x_{k-1}$ such that every contiguous sum lies in $B$. Define \[ \beta_1:=0, \qquad \beta_m:=x_1+x_2+\cdots+x_{m-1} \quad (2\le m\le k). \] Then $\beta_1\le\beta_2\le\cdots\le\beta_k$, and for $1\le i<j\le k$ we have \[ \beta_j-\beta_i=x_i+x_{i+1}+\cdots+x_{j-1}\in B. \] Hence all pairwise differences lie in $B$, so $A$ fails $(P_k)$. This proves the equivalence. $\square$

Corollary 2 (the $k=4$ forbidden pattern)

A set $A\subseteq[0,\infty)$ has $(P_4)$ if and only if its complement \[ B=[0,\infty)\setminus A \] contains no triple $x,y,z\ge 0$ such that \[ x,\ y,\ z,\ x+y,\ y+z,\ x+y+z\in B. \]

Proof. Apply Lemma 1 with $k=4$ and write \[ x_1=x,\quad x_2=y,\quad x_3=z. \] The contiguous sums are then exactly \[ x,\ y,\ z,\ x+y,\ y+z,\ x+y+z. \] $\square$

Theorem 3 (lacunary multi-hole models with $(P_k)$)

Fix an integer $k\ge 3$, a real $L>0$, a hole length $\varepsilon>0$, and a ratio $R>k-1$. Choose a base point $M>0$ such that \[ M>\max\left\{L,\frac{\varepsilon}{k-2},\frac{(k-1)\varepsilon}{R-(k-1)}\right\}. \] Define \[ a_n:=MR^{n-1}\qquad (n\ge 1), \] and let \[ B:=\bigcup_{n\ge 1}(a_n,a_n+\varepsilon), \qquad A:=[0,\infty)\setminus B. \] Then the following hold:

1. $A$ is closed and not dense in $[0,\infty)$.
2. $A$ contains $[0,L]$.
3. $A$ has infinitely many pairwise disjoint complementary intervals.
4. $A$ has property $(P_k)$.
5. The exact closed-interval syndetic constant of $A$ is $\varepsilon$: every closed interval of length $\varepsilon$ meets $A$, and for every $c<\varepsilon$ some closed interval of length $c$ misses $A$.
6. The complement has logarithmic counting growth: \[ \mu(B\cap[0,T])\le \varepsilon\left(1+\frac{\log(T/M)}{\log R}\right) \qquad (T\ge M). \] In particular, \[ \lim_{T\to\infty}\frac{\mu(A\cap[0,T])}{T}=1. \]
7. The complement $B$ has infinite total measure: \[ \mu(B)=\infty. \]

Proof.

$1$ Each interval $(a_n,a_n+\varepsilon)$ is open, so $B$ is open and therefore $A$ is closed. Since every $(a_n,a_n+\varepsilon)$ is a nonempty open interval disjoint from $A$, the set $A$ is not dense.

$2$ Because $a_1=M>L$, the first hole begins to the right of $L$. Hence \[ [0,L]\subseteq [0,M]\subseteq A. \]

$3$ The intervals $(a_n,a_n+\varepsilon)$ are pairwise disjoint because $a_{n+1}=Ra_n>a_n+\varepsilon$. Thus $B$ has infinitely many connected components.

$4$ We use Lemma 1. Assume for contradiction that $A$ fails $(P_k)$. Then there exist nonnegative reals \[ x_1,\dots,x_{k-1} \] all of whose contiguous sums lie in $B$. In particular, each singleton sum $x_i$ lies in $B$, so for each $i$ there exists an index $n_i$ with \[ a_{n_i}<x_i<a_{n_i}+\varepsilon. \] Let \[ n:=\max\{n_1,\dots,n_{k-1}\}. \] Then at least one of the $x_i$ is $>a_n$, while every $x_i$ is $<a_n+\varepsilon$. Therefore the total sum \[ S:=x_1+x_2+\cdots+x_{k-1} \] satisfies \[ S>a_n+(k-2)M>a_n+\varepsilon \] by the choice $M>\varepsilon/(k-2)$, and also \[ S<(k-1)(a_n+\varepsilon)<Ra_n=a_{n+1} \] by the choice $M>\frac{(k-1)\varepsilon}{R-(k-1)}$ (since $a_n\ge M$). Hence \[ a_n+\varepsilon < S < a_{n+1}. \] But the set $B$ has no points in the interval $(a_n+\varepsilon,a_{n+1})$, because its components are exactly the holes $(a_m,a_m+\varepsilon)$. Thus $S\notin B$, contradicting the assumption that all contiguous sums lie in $B$. Therefore $A$ has $(P_k)$.

$5$ First we show that every closed interval of length $\varepsilon$ meets $A$. Suppose not. Then some closed interval $I=[u,u+\varepsilon]$ is contained in $B$. Since $I$ is connected and the connected components of $B$ are exactly the disjoint open intervals $(a_n,a_n+\varepsilon)$, the interval $I$ must be contained in one component, say $(a_n,a_n+\varepsilon)$. But a closed interval of length $\varepsilon$ cannot be contained in an open interval of the same length. Contradiction. So every closed interval of length $\varepsilon$ meets $A$.

Now let $0<c<\varepsilon$. Then the closed interval \[ \left[a_1+\frac{\varepsilon-c}{2},\ a_1+\frac{\varepsilon+c}{2}\right] \] has length $c$ and lies entirely inside the hole $(a_1,a_1+\varepsilon)$, so it is disjoint from $A$. Hence no smaller positive constant works. Therefore the exact closed-interval syndetic constant is $\varepsilon$.

$6$ For $T\ge M$, let $N(T)$ denote the number of holes whose left endpoint is $<T$. Since $a_n=MR^{n-1}$, the inequality $a_n<T$ implies \[ n-1<\frac{\log(T/M)}{\log R}. \] Therefore \[ N(T)\le 1+\frac{\log(T/M)}{\log R}. \] Each hole contributes at most $\varepsilon$ of measure to $B\cap[0,T]$, so \[ \mu(B\cap[0,T])\le \varepsilon N(T) \le \varepsilon\left(1+\frac{\log(T/M)}{\log R}\right). \] Hence \[ \mu(A\cap[0,T])=T-\mu(B\cap[0,T])=T-O(\log T), \] which implies \[ \frac{\mu(A\cap[0,T])}{T}\to 1. \]

$7$ The holes are pairwise disjoint and each has measure $\varepsilon$, so \[ \mu(B)=\sum_{n\ge 1}\mu((a_n,a_n+\varepsilon)) =\sum_{n\ge 1}\varepsilon =\infty. \] This completes the proof. $\square$

Corollary 4 (the case relevant to the prime-gap problem)

For every $L>0$ and every $\varepsilon>0$, there exists a closed non-dense set $A\subseteq[0,\infty)$ such that:

1. $[0,L]\subseteq A$;
2. $A$ has property $(P_4)$;
3. the exact closed-interval syndetic constant of $A$ is $\varepsilon$;
4. $A$ has infinitely many complementary intervals;
5. $\mu([0,\infty)\setminus A)=\infty$;
6. $\displaystyle \lim_{T\to\infty}\frac{\mu(A\cap[0,T])}{T}=1$.

Proof. Apply Theorem 3 with $k=4$ and, for instance, $R=4$. Then it is enough to choose \[ M>\max\{L,\varepsilon/2,3\varepsilon\}=\max\{L,3\varepsilon\}. \] All the claimed properties follow immediately. $\square$

Corollary 5 (a stronger no-go theorem)

There do not exist constants $L_0>0$ and $\varepsilon_0>0$ with the following property:

every closed set $A\subseteq[0,\infty)$ satisfying

• $[0,L_0]\subseteq A$,
• $A$ has $(P_4)$,
• the exact closed-interval syndetic constant of $A$ is at most $\varepsilon_0$,
• $[0,\infty)\setminus A$ has infinitely many connected components,
• $\displaystyle \lim_{T\to\infty}\frac{\mu(A\cap[0,T])}{T}=1$,

must be dense in $[0,\infty)$.

Proof. Fix arbitrary $L_0,\varepsilon_0>0$. Apply Corollary 4 with $L=L_0$ and $\varepsilon=\varepsilon_0/2$. The resulting set $A$ is closed, not dense, contains $[0,L_0]$, has $(P_4)$, has exact closed-interval syndetic constant $\varepsilon_0/2<\varepsilon_0$, has infinitely many complementary intervals, and has asymptotic density $1$. So it satisfies all the displayed hypotheses but is not dense, contradiction. $\square$

Interpretation. Round--5 had already shown that one-hole countermodels destroy any proof strategy based only on closedness, $(P_4)$, syndeticity, and large measure. Round--6 strengthens this decisively: even infinitely many holes, infinite complement measure, and asymptotic density $1$ do not force density. Thus any proof of Erdős' conjecture for the genuine prime-gap limit-point set must exploit arithmetic information that is invisible to these coarse geometric and additive-combinatorial invariants.

ADVERSARIAL VERIFICATION

• Boundary case in Lemma 1. The proof allows $x_i=0$, which is necessary because equal consecutive $\beta$'s are permitted in the definition of $(P_k)$. No positivity assumption was smuggled in.

• Are the components of $B$ really the single holes $(a_n,a_n+\varepsilon)$? Yes. The intervals are pairwise disjoint because $a_{n+1}=Ra_n>a_n+\varepsilon$ by construction. Hence connected components of $B$ are exactly those open intervals.

• Upper-bound check in Theorem 3$4$. The key estimate is \[ S<(k-1)(a_n+\varepsilon)<Ra_n=a_{n+1}. \] This follows from \[ (k-1)\varepsilon < (R-(k-1))a_n, \] which holds because $a_n\ge M$ and \[ M>\frac{(k-1)\varepsilon}{R-(k-1)}. \] No hidden dependence on $n$ remains.

• Lower-bound check in Theorem 3$4$. We used \[ S>a_n+(k-2)M>a_n+\varepsilon. \] This is valid because one summand exceeds $a_n$ and the other $k-2$ summands each exceed $M$, since every singleton sum $x_i$ lies in some hole and therefore is $>a_1=M$.

• Exact syndetic constant. The proof is with closed intervals, matching the convention used in earlier rounds. For open-interval syndeticity one would take any constant $>\varepsilon$ instead. This does not affect the obstruction statements.

• Interaction with Round--5. Round--5 handled one-gap models; Round--6 does not reprove that result but extends the obstruction to infinitely many holes. The two are consistent: the new family simply shows that even finite-component assumptions are unnecessary for countermodels.

• Relevance to the actual set $S$. The constructed sets are model sets, not claims about prime gaps. Their purpose is logical: to show that currently known coarse properties of $S$ are insufficient to imply density.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round--6 makes two substantive advances beyond Round--5.

1. It proves a new complement-side structural characterization of $(P_k)$. For $k=4$, this becomes the exact forbidden additive pattern \[ x,\ y,\ z,\ x+y,\ y+z,\ x+y+z. \] This is a genuinely new tool for constructing and analyzing model sets.

2. It constructs infinitely many-hole countermodels with the same coarse properties one might hope to use in proving density: closedness, $(P_4)$, arbitrarily long initial segment containment, arbitrarily small exact syndetic constant, and asymptotic density $1$. In fact these countermodels may also have infinite complement measure and infinitely many complementary components.

So the gap isolated after Round--5 is now closed: density cannot be deduced even after excluding one-hole or finite-measure-complement pathologies. A successful proof of Erdős' conjecture must therefore use arithmetic structure beyond all presently known coarse invariants treated in Rounds 3--6.

COMPLETION ESTIMATE

COMPLETION: 74%

REFERENCES

• T. F. Bloom, Erdős Problem #5, erdosproblems.com/5.
• W. D. Banks, T. Freiberg, J. Maynard, On limit points of the sequence of normalized prime gaps, arXiv:1404.5094v2.
• J. Merikoski, Limit points of normalized prime gaps, arXiv:1811.03008v3.
• J. Pintz, On the distribution of gaps between consecutive primes, arXiv:1407.2213.

Erdős Problem #5: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

For primes $p_n$, is the set $S$ of finite limit points of $(p_{n+1}-p_n)/\log p_n$ all of $[0,\infty)$?

All supplied versions considered: 5.tex. The source constructs infinitely many complementary holes satisfying the four-point difference property $(P_4)$. Its particular model actually satisfies the stronger three-point property. This improves the logical audit, not the prime-gap theorem.

2) WORK

Fix $\epsilon>0$, $M>\epsilon$ and $R>2+2\epsilon/M$. Put \[ B=\bigcup_{j\ge0}(MR^j,MR^j+\epsilon),\qquad A=[0,\infty)\setminus B. \] I claim $B$ is sum-free: $(B+B)\cap B=\varnothing$. If $x,y\in B$, and $a=MR^j$ is the larger of their two hole origins, then \[ a+\epsilon<a+M<x+y<2(a+\epsilon)<Ra. \] There are no holes between $a+\epsilon$ and $Ra$, which proves the claim. Given $\beta_1\le\beta_2\le\beta_3$, if both consecutive differences are in $B$, their sum is in $A$; if one is outside $B$ there is already a difference in $A$. Hence $A$ has $(P_3)$, and therefore $(P_k)$ for every $k\ge3$.

In particular the supplied choice $R=4$, $M>3\epsilon$ has this stronger property. Nevertheless $A$ is closed and omits every nonempty open interval $(MR^j,MR^j+\epsilon)$. It is not dense. Its complement has infinite measure and only $O(\log T)$ measure below $T$, by counting the geometrically spaced origins. Thus even adding $(P_3)$ to the listed geometric hypotheses cannot force density for arbitrary model sets.

3) ADVERSARIAL VERIFICATION

The argument never asserts that the actual prime-gap set satisfies $(P_3)$. A model satisfying a stronger property still refutes the corresponding abstract implication, but cannot refute a conjecture about primes. All inequalities are strict because hole endpoints belong to $A$.

4) FINAL

UNRESOLVED. No new normalized prime gap is produced. An arithmetic input implying every prescribed interval meets $S$ is still absent; extending the countermodel class is not substantial completion of that requirement.

5) SOURCES CHECKED

Read 5.tex in full and checked Merikoski, Limit points of normalized prime gaps, https://arxiv.org/abs/1811.03008 $2026-09-23$: its abstract confirms the $T/3$ measure bound and bounded gaps. The model proof above is independent of that theorem.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 5%

Round 2 Objective

Path chosen: (C) Obstruction / correction via rigorous reduction.

Building strictly on the Round--1 results, the objective of Round--2 is not to restart the investigation, but to sharpen and reduce the conjecture to its strongest known and most standard formulation. The key advance is to show that the normalization used in the problem, \[ g_n=\frac{p_{n+1}-p_n}{\log n}, \] has exactly the same set of limit points as the classical normalization \[ \frac{p_{n+1}-p_n}{\log p_n} \] used in the modern literature. This allows all known deep results to be transferred verbatim to the present setting and isolates a single precise analytic obstruction whose resolution would settle the conjecture.

Round--1 Foundation Used

We rely on the following Round--1 results without reproving them:

• Lemma 1 (Closedness). The limit--point set $S$ of $(g_n)$ is a closed subset of $[0,\infty]$.
• Lemma 2. $0\in S \iff \liminf g_n=0$ and $\infty\in S \iff \limsup g_n=\infty$.
• Lemma 3. If $S$ is dense in $[0,\infty]$, then $S=[0,\infty]$.
• The definition of $g_n$ and the numerical sanity checks up to $n\le 10^5$.

New Insight / Tool Introduced in Round--2

The new ingredient is a general normalization invariance lemma: multiplying a sequence by a factor $r_n\to 1$ does not change its set of limit points (including $\infty$). Applying this to \[ r_n=\frac{\log p_n}{\log n}, \] which tends to $1$, shows that the limit--point set defined using $\log n$ coincides exactly with the classical limit--point set defined using $\log p_n$. This reduction allows us to import the strongest known theorems (Merikoski; Banks--Freiberg--Maynard) directly into the present problem.

Attack Plan

The remaining Round--1 gap is density of $S$. The Round--2 plan is:

1. Prove that rescaling by $r_n\to 1$ preserves limit points.
2. Prove $\log p_n/\log n\to 1$ using explicit bounds for $p_n$.
3. Deduce that the present set $S$ equals the classical limit--point set $L$.
4. Import the strongest known results about $L$ (large measure, bounded gaps).
5. Isolate the precise analytic obstruction whose improvement would imply $S=[0,\infty]$.

Work

Rescaling Lemma

Let $(a_n)$ be a sequence in $[0,\infty]$ and let $(r_n)$ be positive with $r_n\to 1$. Define $b_n=a_n r_n$. Then $(a_n)$ and $(b_n)$ have the same set of limit points in $[0,\infty]$.

Proof.

If $a_{n_k}\to x<\infty$, then $b_{n_k}=a_{n_k}r_{n_k}\to x$. Conversely, if $b_{n_k}\to x<\infty$, then $a_{n_k}=b_{n_k}/r_{n_k}\to x$. For $\infty$, since $r_n\to 1$, eventually $1/2\le r_n\le 2$, so divergence of one sequence forces divergence of the other. $\square$

Asymptotics of $\log p_n$

Proof.

Since $p_n\ge n+1$, we have $\log p_n/\log n\ge \log(n+1)/\log n\to 1$. Explicit bounds of Dusart give \[ p_n\le n(\log n+\log\log n+O(1)), \] hence $\log p_n\le \log n+o(\log n)$, which implies the limsup is at most $1$. $\square$

Equality of Limit--Point Sets

Define \[ g_n=\frac{p_{n+1}-p_n}{\log n},\qquad h_n=\frac{p_{n+1}-p_n}{\log p_n}. \] Let $S$ and $L$ be the respective limit--point sets.

Proof.

We have $g_n=h_n r_n$ with $r_n=\log p_n/\log n\to 1$. By the rescaling lemma, the two sequences have identical limit--point sets. $\square$

Imported Consequences

Identifying $S=L$, we obtain immediately:

• (Merikoski) For all $T\ge 0$, $\mu(S\cap[0,T])\ge T/3$, and $S$ is syndetic (has bounded gaps).
• (Banks--Freiberg--Maynard) Quantitative difference--set properties and positive lower density of $S$.

New Corollary

Proof.

Merikoski's four--point difference theorem applied to $\{0,x,2x,3x\}$ implies that one of the differences $x,2x,3x$ lies in $L=S$. $\square$

Adversarial Verification

• Finite truncations do not affect limit points.
• The rescaling argument handles $\infty$ explicitly.
• No unproven hypotheses (e.g. RH, EH) are used in the reduction.
• Closedness plus bounded gaps does not imply density, so no logical overreach occurs.

Final Status

UNRESOLVED (BUT STRICTLY ADVANCED).

The conjecture $S=[0,\infty]$ is reduced exactly to the standard formulation for normalized prime gaps. All strongest known partial results now apply directly to the original problem, and a precise analytic obstruction is isolated: improving the key sieve bound in Merikoski's framework would settle the conjecture.

Completion Estimate

COMPLETION: 45%

References

• P. Dusart, Estimates of some functions over primes without R.H.
• W. D. Banks, T. Freiberg, J. Maynard, On limit points of the sequence of normalized prime gaps.
• J. Merikoski, Limit points of normalized prime gaps.

ROUND-3 OBJECTIVE

Path chosen: (C) Obstruction / correction (strategy-elimination) + refined necessary form of counterexamples.

Round--2 reduced the original normalization $\log n$ to the standard $\log p_n$ normalization (showing the limit-point sets coincide) and imported the strongest known measure/syndeticity statements for the classical limit-point set. The Round--1 core gap remains: prove topological density of $S$ (hence $S=[0,\infty]$ by closedness).

In Round--3 we pursue a rigorous gap-closure obstruction: we prove that the currently known structural properties of $S$ (closedness, syndeticity, positive-measure lower bounds, and the "$k$-point difference" property) do not logically imply density. Concretely, we build a closed, syndetic, high-measure set satisfying these same properties while still missing an open interval. This rules out an entire class of argument templates and sharpens the minimal shape any counterexample interval must have.

ROUND-1/2 FOUNDATION USED

We use (and do not reprove) the following vetted Round--1/2 results:

• (Round--1 Lemma 1) $S$ is closed in $[0,\infty]$.
• (Round--1 Lemma 3) If $S$ is dense in $[0,\infty]$, then $S=[0,\infty]$.
• (Round--2 Reduction) The limit-point set for $\frac{p_{n+1}-p_n}{\log n}$ equals the classical limit-point set for $\frac{p_{n+1}-p_n}{\log p_n}$. Thus we may freely import theorems stated for the latter.
• (Imported in Round--2) Merikoski: $\mu(S\cap[0,T])\ge T/3$ for all $T\ge 0$ and $S$ is syndetic (bounded additive gaps between limit points).
• (Imported in Round--2) Banks--Freiberg--Maynard: the $k=9$ "difference-set" property (for any $9$ reals, some pairwise difference lies in $S$).

NEW INSIGHT / TOOL (ROUND-3)

New tool: a quantitative relative-gap lemma for sets with the $k$-point difference property. It implies any open interval missing from such a set must have bounded multiplicative ratio, and (in the closed case) forces intersection with every interval $[a,(k-1)a]$.

New obstruction: a concrete closed, syndetic, high-measure counterexample set satisfying all currently known structural properties of $S$ but still not dense. This demonstrates that density cannot be deduced from these properties alone and therefore identifies a genuine missing ingredient needed to resolve Erdős' conjecture.

ATTACK PLAN (ROUND-3)

1. Use the known $k$-point difference property (for $k=4$ and/or $k=9$) to prove a new lemma: if a set $A$ has this property, then it cannot omit an interval $(a,b)$ with $b>(k-1)a$.
2. Combine with closedness to sharpen: a closed $A$ with the property must intersect every $[a,(k-1)a]$.
3. Construct an explicit closed set $A\subset[0,\infty)$ that is syndetic, satisfies $\mu(A\cap[0,T])\ge T/3$ for all $T$, and satisfies the same $k$-point difference property as $S$, yet is not dense.
4. Conclude: any proof of $S=[0,\infty]$ must exploit additional arithmetic structure beyond the currently known abstract properties; and any counterexample interval for $S$ must obey new quantitative restrictions.

WORK (ROUND-3)

Imported theorem: $S$ contains an initial interval $[0,c]$ (ineffective $c>0$)

In the classical normalization, it is known (via Zhang's bounded gaps) that there exists an ineffective constant $c>0$ such that $[0,c]\subseteq S$. Equivalently, every real number in $[0,c]$ is a limit point of normalized prime gaps.

This is strictly stronger than $0\in S$: it gives nonempty interior near $0$.

Definition: the $k$-point difference property

Let $A\subseteq[0,\infty)$. For an integer $k\ge 2$, say that $A$ has property $(P_k)$ if for every real numbers \[ \beta_1\le\beta_2\le\cdots\le\beta_k \] we have \[ A\cap\{\beta_j-\beta_i:1\le i<j\le k\}\neq\emptyset. \] By Merikoski's Theorem 1, $S$ has $(P_4)$; by Banks--Freiberg--Maynard, $S$ has $(P_9)$.

New Lemma: relative-gap constraint from $(P_k)$

Proof.

Assume for contradiction that $(a,b)\cap A=\emptyset$ with $b>(k-1)a$. Choose $\varepsilon>0$ such that \[ a+\varepsilon<\frac{b}{k-1}, \] which is possible exactly because $b>(k-1)a$.

Define $k$ points by \[ \beta_i:=(i-1)(a+\varepsilon),\qquad i=1,2,\dots,k. \] Then every difference has the form \[ \beta_j-\beta_i=(j-i)(a+\varepsilon)\in\{(a+\varepsilon),2(a+\varepsilon),\dots,(k-1)(a+\varepsilon)\}. \] Each such number lies in $(a,b)$ since $a+\varepsilon>a$ and $(k-1)(a+\varepsilon)<b$. Hence \[ \{\beta_j-\beta_i:1\le i<j\le k\}\subset (a,b), \] so it is disjoint from $A$, contradicting $(P_k)$. $\square$

Corollary: closed + $(P_k)$ forces multiplicative syndeticity

Proof.

Fix $a>0$ and suppose $A\cap[a,(k-1)a]=\emptyset$. Since $A$ is closed and $[a,(k-1)a]$ is compact, the distance \[ d:=\mathrm{dist}\left(A,\,[a,(k-1)a]\right) \] is strictly positive. Put $\delta:=\min(d/2,a/2)$ so that $a-\delta>0$ and \[ (a-\delta,\,(k-1)a+\delta)\cap A=\emptyset. \] But then \[ (k-1)a+\delta > (k-1)(a-\delta) \] (since the difference is $k\delta>0$), hence the interval $(a-\delta,(k-1)a+\delta)$ has right endpoint $>(k-1)$ times its left endpoint. This contradicts the relative-gap lemma. $\square$

New obstruction example: known structural properties do not imply density

We now show that the package of currently known abstract properties of $S$ is insufficient to deduce density.

Proof.

$1$ Closedness is immediate.

$2$ Let $T\ge 0$. If $T\le 1$, then $[T,T+2]$ intersects $[0,1]\subset A$. If $1<T<3$, then $3\in[T,T+2]\cap A$. If $T\ge 3$, then $[T,T+2]\subset[3,\infty)\subset A$. Hence every interval of length $2$ intersects $A$.

$3$ Compute $\mu(A\cap[0,T])$: \[ \mu(A\cap[0,T])= \begin{cases} T, & 0\le T\le 1,\\ 1, & 1<T\le 3,\\ 1+(T-3)=T-2, & T\ge 3. \end{cases} \] For $0\le T\le 1$, we have $T\ge T/3$. For $1<T\le 3$, we have $1\ge T/3$. For $T\ge 3$, we have $T-2\ge T/3$ since $2T/3\ge 2$.

$4$ To prove $(P_4)$, let $\beta_1\le\beta_2\le\beta_3\le\beta_4$. If any consecutive difference satisfies $\beta_{i+1}-\beta_i\le 1$, then that difference lies in $[0,1]\subset A$ and we are done. Otherwise all three consecutive differences exceed $1$, so \[ \beta_4-\beta_1=(\beta_2-\beta_1)+(\beta_3-\beta_2)+(\beta_4-\beta_3)>3, \] hence $\beta_4-\beta_1\in[3,\infty)\subset A$, and we are done. Thus $(P_4)$ holds. The proof for $(P_9)$ is identical: if no consecutive difference is $\le 1$, then $\beta_9-\beta_1>8$, hence belongs to $[3,\infty)\subset A$.

$5$ $A$ is not dense because $(1,3)\cap A=\emptyset$. $\square$

Round--3 consequence: what a counterexample interval for $S$ must look like

Assume (for contradiction) that $S\neq[0,\infty]$. By Round--1 Lemma 1, there exists a nonempty open interval $(a,b)\subset[0,\infty)$ with $(a,b)\cap S=\emptyset$.

Combining:

• the imported fact that $[0,c]\subseteq S$ for some $c>0$ (ineffective), and
• the corollary from $(P_4)$ (relative-gap obstruction), and
• syndeticity of $S$ (bounded additive gaps),

we obtain the following necessary constraints on any such missing interval $(a,b)$:

1. $a\ge c$ (since $(0,c)$ is fully contained in $S$).
2. $b\le 3a$ (since $(P_4)$ forbids gaps with multiplicative ratio $>3$).
3. $b-a\le C_S$ for some absolute (ineffective) syndeticity constant $C_S$ (since $S$ intersects every interval of length $C_S$).

Thus any failure of density must come from a short additive gap occurring above the initial interval $[0,c]$, and with bounded multiplicative ratio.

Crucial meta-point: Proposition above shows these constraints and the known abstract properties are far from sufficient to force density; any genuine proof of Erdős' conjecture must use arithmetic structure beyond closedness + syndeticity + measure bounds + $(P_k)$.

ADVERSARIAL VERIFICATION

• Quantifier check for $(P_k)$. The proofs use only that $\beta_1\le\cdots\le\beta_k$; differences are automatically nonnegative, consistent with $A\subset[0,\infty)$.
• Edge case $a\to 0$ in the corollary. We ensured positivity of the left endpoint by choosing $\delta=\min(d/2,a/2)$ so $a-\delta>0$.
• Tightness of the relative-gap lemma. The lemma excludes $b>(k-1)a$ but permits $b=(k-1)a$. The example $A=[0,1]\cup[3,\infty)$ has an actual missing interval $(1,3)$ with ratio exactly $3$, confirming the bound is sharp in general.
• Interaction with known results for $S$. Since the example set $A$ satisfies the same package of known properties as $S$ (closed, syndetic, $\mu(\cdot)\ge T/3$, $(P_4)$ and $(P_9)$), it rigorously demonstrates that these properties alone cannot imply density.
• No circularity. The example $A$ is purely combinatorial; it does not depend on prime gaps. It is used only to refute a class of deduction strategies, not to claim anything about the actual $S$ beyond logical independence.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

New Round--3 contributions beyond Round--2:

1. Imported (via the standard literature) the stronger fact that $S$ contains a full initial interval $[0,c]$ for some (ineffective) $c>0.
2. Proved a new relative-gap lemma: if a set has the $k$-point difference property $(P_k)$, it cannot omit an interval $(a,b)$ with $b>(k-1)a$; for closed sets this forces intersection with every $[a,(k-1)a]$.
3. Gave an explicit closed, syndetic, high-measure set $A=[0,1]\cup[3,\infty)$ satisfying $(P_4)$ and $(P_9)$ but not dense, showing that the current package of abstract properties known for $S$ is insufficient to deduce Erdős' conjecture.
4. Sharpened the necessary form of any counterexample interval for $S$: it must start above $c$ and satisfy $b\le 3a$ and $b-a\le C_S$.

These results do not resolve density, but they rule out a broad family of "purely topological/combinatorial" proof attempts and sharpen the target for any genuine analytic advance.

COMPLETION ESTIMATE

COMPLETION: 55%

REFERENCES

• W. D. Banks, T. Freiberg, J. Maynard, On limit points of the sequence of normalized prime gaps.
• J. Merikoski, Limit points of normalized prime gaps.

ROUND-4 OBJECTIVE

Path chosen: (C) Obstruction / strategy-elimination + sharpness.

Round--3 isolated a genuine obstruction: from the currently known abstract properties of the limit-point set $S$ (closedness, syndeticity, positive measure, and the $k$-point difference property $(P_k)$), one cannot deduce density of $S$. The Round--3 example $A=[0,1]\cup[3,\infty)$ already showed that these properties do not imply $S=[0,\infty)$.

In Round--4 we strengthen this obstruction in two mathematically substantive ways:

1. We prove a sharpness theorem showing that the Round--3 multiplicative gap constraint $b\le 3a$ is essentially optimal: for any interval $(a,b)$ with $0<a<b\le 3a$ there exists a closed, syndetic set satisfying $(P_4)$ and the measure lower bound $\mu(\cdot)\ge T/3$, yet missing $(a,b)$.
2. We build an infinite family of such counter-model sets with missing intervals occurring arbitrarily far out while simultaneously containing an arbitrarily long initial segment $[0,L]$ and even having very small syndetic constant. This rules out an entire class of proof strategies that try to deduce density from the known "global" structural properties of $S$.

These results do not resolve Erdős' conjecture for the actual set $S$, but they rigorously show that any successful proof must use additional arithmetic structure beyond $(P_4)$-type combinatorics, measure bounds, and syndeticity.

ROUND-3 FOUNDATION USED

We rely on the following Round--3 results (treated as vetted progress):

• (R1 Lemma 1) The limit-point set $S$ is closed in $[0,\infty]$.
• (R3 Definition) $(P_k)$: for all $\beta_1\le\cdots\le\beta_k$, the difference set $\{\beta_j-\beta_i:1\le i<j\le k\}$ meets $S$.
• (R3 Relative-gap lemma) If a set $A\subset[0,\infty)$ has $(P_k)$ and $(a,b)\cap A=\emptyset$ with $a>0$, then $b\le (k-1)a$. In particular, for $(P_4)$ one must have $b\le 3a$.
• (R3 Corollary) If $A$ is closed and has $(P_k)$, then $A\cap[a,(k-1)a]\neq\emptyset$ for every $a>0$.
• (Imported earlier) For the actual prime-gap limit-point set $S$, Merikoski proves $(P_4)$, $\mu(S\cap[0,T])\ge T/3$ for all $T\ge 0$, and $S$ is syndetic; and also cites Pintz that $[0,c]\subseteq S$ for some ineffective $c>0$.

NEW INSIGHT / TOOL (ROUND-4)

New tool: a sharpness construction for $(P_k)$.

We show that the Round--3 multiplicative gap constraint is best possible in the following strong sense: whenever $0<a<b\le (k-1)a$, the simple two-block set \[ A_{a,b}:=[0,a]\cup[b,\infty) \] is closed, syndetic, satisfies $(P_k)$, and already saturates the general measure lower bound $\mu(A\cap[0,T])\ge T/(k-1)$ at $T=b$ when $b=(k-1)a$.

In particular, for $k=4$ we obtain a large family of counter-models that satisfy the same abstract properties currently proved for $S$, while still missing an arbitrary prescribed open interval $(a,b)$ with $b\le 3a$. This strictly strengthens the Round--3 strategy-elimination example.

ATTACK PLAN (ROUND-4)

1. Prove the sharpness construction: $A_{a,b}=[0,a]\cup[b,\infty)$ satisfies $(P_k)$ whenever $b\le (k-1)a$.
2. Verify that the same $A_{a,b}$ is closed, syndetic, and satisfies the measure bound $\mu(A_{a,b}\cap[0,T])\ge T/(k-1)$ for all $T\ge 0$ (hence $\ge T/3$ for $k=4$).
3. Specialize to $k=4$ and exhibit an infinite family with:
   • missing intervals located arbitrarily far out (large $a$),
   • arbitrarily long initial segment containment $[0,L]\subset A_{a,b}$,
   • and even very small syndetic constants (e.g. $1$).
4. Conclude that density of $S$ cannot be deduced from the presently known abstract properties; a proof of Erdős' conjecture must exploit additional arithmetic structure.

WORK (ROUND-4)

Reminder: $(P_k)$

For $k\ge 2$, a set $A\subseteq[0,\infty)$ has property $(P_k)$ if for every choice of real numbers $\beta_1\le\cdots\le\beta_k$ we have \[ A\cap\{\beta_j-\beta_i:1\le i<j\le k\}\neq\emptyset. \] (As noted in Round--3, $(P_4)$ implies $(P_k)$ for all $k\ge 4$ by restriction to any four points.)

Sharpness construction for $(P_k)$

Proof.

$1$ Closedness is clear since $A_{a,b}$ is a union of closed intervals. Non-density holds since $(a,b)\cap A_{a,b}=\emptyset$.

$2$ Let $C:=b-a$. We claim every closed interval $[x,x+C]$ intersects $A_{a,b}$. If $x\le a$, then $[x,x+C]$ intersects $[0,a]\subset A_{a,b}$. If $x>a$, then $x+C>x+(b-a)>b$, hence $[x,x+C]$ intersects $[b,\infty)\subset A_{a,b}$. Thus $A_{a,b}$ is syndetic.

$3$ Let $\beta_1\le\cdots\le\beta_k$. If some consecutive gap $\beta_{i+1}-\beta_i\le a$, then this difference lies in $[0,a]\subset A_{a,b}$. Otherwise every consecutive gap satisfies $\beta_{i+1}-\beta_i>a$, so \[ \beta_k-\beta_1=\sum_{i=1}^{k-1}(\beta_{i+1}-\beta_i)>(k-1)a\ge b, \] hence $\beta_k-\beta_1\in[b,\infty)\subset A_{a,b}$. In all cases, some pairwise difference lies in $A_{a,b}$, i.e. $(P_k)$ holds.

$4$ We compute $\mu(A_{a,b}\cap[0,T])$ explicitly:

\[ \mu(A_{a,b}\cap[0,T])= \begin{cases} T, & 0\le T\le a,\\ a, & a<T<b,\\ a+(T-b)=T-(b-a), & T\ge b. \end{cases} \]

If $0\le T\le a$, then $\mu=T\ge T/(k-1)$. If $a<T<b$, then $\mu=a\ge b/(k-1)\ge T/(k-1)$ since $T<b\le (k-1)a$. If $T\ge b$, then \[ \mu=T-(b-a)\ \ge\ T-\left(b-\frac{b}{k-1}\right)=\frac{T}{k-1}, \] because $a\ge b/(k-1)$. This proves the measure bound for all $T\ge 0$. $\square$

Consequences for $k=4$: optimality of the Round--3 constraint

Specializing Theorem Theorem (Sharpness of the relative-gap constraint) to $k=4$ yields:

Together with the Round--3 relative-gap lemma (necessity of $b\le 3a$ for any set with $(P_4)$), this shows that the multiplicative restriction $b\le 3a$ is optimal: it cannot be improved purely from $(P_4)$.

A strengthened strategy-elimination family: far-out holes + long initial segment

Fix any $L>0$ and any $M>L$ with $M\ge \tfrac12$. Set $a:=M$ and $b:=M+1$ (so $b\le 3a$ holds for $M\ge \tfrac12$) and define \[ A_M := A_{M,M+1}=[0,M]\cup[M+1,\infty)= [0,\infty)\setminus(M,M+1). \] Then:

• $[0,L]\subseteq A_M$ (indeed, $[0,M]\subseteq A_M$), so $A_M$ contains an arbitrarily long initial interval.
• $A_M$ is syndetic with constant $1$ (every closed interval of length $1$ meets $A_M$).
• By Theorem Theorem (Sharpness of the relative-gap constraint) (with $k=4$), $A_M$ satisfies $(P_4)$ and $\mu(A_M\cap[0,T])\ge T/3$ for all $T$.
• $A_M$ is not dense since it misses the open interval $(M,M+1)$, and this missing interval can be placed arbitrarily far out by taking $M$ large.

Conclusion. Even if one strengthens the abstract hypotheses drastically --- closedness, $(P_4)$, the uniform measure bound $\mu(\cdot)\ge T/3$ for all $T$, syndeticity with tiny constant, and containment of an arbitrarily long initial segment --- density still does not follow. Hence, any proof of Erdős' conjecture for the genuine prime-gap set $S$ must use additional arithmetic structure beyond these global properties.

ADVERSARIAL VERIFICATION

• Boundary case $b=(k-1)a$. In the $(P_k)$ proof we use $\beta_k-\beta_1>(k-1)a$, hence $\beta_k-\beta_1>b$ even when $b=(k-1)a$, so membership in $[b,\infty)$ is valid.
• Syndeticity convention (open vs. closed intervals). The proof uses closed intervals $[x,x+C]$. If one uses open intervals $(x,x+C)$, then one should take any $C'>b-a$ instead; this does not affect the strategy-elimination conclusions.
• Measure bound check in the regime $a<T<b$. The inequality reduces to $a\ge T/(k-1)$ for all $T<b$, which is guaranteed by $b\le (k-1)a$ and is tight at $T=b$ when $b=(k-1)a$.
• Interaction with Round--3 relative-gap lemma. Our construction matches the Round--3 necessary condition $b\le (k-1)a$, showing it is sharp; there is no contradiction.
• Relevance to the actual $S$. These sets $A_{a,b}$ are model sets sharing the same abstract properties currently proved for $S$. They do not claim anything about primes, but they rigorously show why these properties alone cannot settle the conjecture.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round--4 establishes sharpness and a stronger strategy-elimination principle:

1. The Round--3 multiplicative restriction on missing intervals for sets with $(P_4)$, namely $b\le 3a$, is optimal: any open interval $(a,b)$ with $b\le 3a$ can be removed from a closed syndetic set still satisfying $(P_4)$ and $\mu(\cdot)\ge T/3$ for all $T$.
2. There exist counter-model sets with arbitrarily long initial intervals and arbitrarily far-out missing intervals, even with syndetic constant $1$ and the full uniform measure lower bound $T/3$.

Thus any proof of Erdős' conjecture for the genuine limit-point set of normalized prime gaps must use substantially more arithmetic information than the currently known global structural properties.

COMPLETION ESTIMATE

COMPLETION: 60%

REFERENCES

• W. D. Banks, T. Freiberg, J. Maynard, On limit points of the sequence of normalized prime gaps, arXiv:1404.5094v2.
• J. Merikoski, Limit points of normalized prime gaps, arXiv:1811.03008v3.
• J. Pintz, On the distribution of gaps between consecutive primes, arXiv:1407.2213 (for the interval $[0,c]\subseteq L$ result cited by Merikoski).