we-solve (beta)

Math-6011-A2 · revision r1 · on Math-6011

gpt pro 5.2 attempt on Math-6011

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #949
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE/CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. VERIFICATION
  7. FINAL

Erdős Problem #949

FORMAL RESTATEMENT

Let $\mathfrak c:=|\mathbb R|$.

Literal statement (as written). Given a set $S\subset\mathbb R$ such that there do not exist $a,b,c\in S$ with $a+b=c$, must there exist a set $A\subseteq\mathbb R\setminus S$ with $|A|=\mathfrak c$ such that \[ A+A:=\{x+y:x,y\in A\}\subseteq \mathbb R\setminus S? \] (Edge cases: the condition rules out $0\in S$ because $0+0=0$ would be a solution.)

Variant mentioned in the problem text. Assume instead that $S$ is Sidon in $(\mathbb R,+)$, meaning: for any $s_1,s_2,s_3,s_4\in S$, if $s_1+s_2=s_3+s_4$ then $\{s_1,s_2\}=\{s_3,s_4\}$. Must there exist $A\subseteq\mathbb R\setminus S$ with $|A|=\mathfrak c$ and $A+A\subseteq\mathbb R\setminus S$?

QUICK LITERATURE/CONTEXT CHECK

The extracted problem text records: (i) the general sum-free version above (no $a+b=c$ in $S$) is open on the ErdosProblems site; (ii) Erdős suggested strengthening to Sidon sets; (iii) a positive proof for the Sidon variant was posted in the comments (attributed there to "AlphaProof"). I will not use any external results beyond what is stated in the problem text; I reproduce full proofs below.

ATTACK PLAN

  • Proof track (partial): Prove the statement when $|S|<\mathfrak c$ by taking a maximal $A$ (Zorn) with $A\cap S=\emptyset$ and $(A+A)\cap S=\emptyset$, and use maximality plus a cardinality bound to force $|A|=\mathfrak c$.
  • Proof track (Sidon variant): If $S$ is Sidon and $|S|=\mathfrak c$, give an explicit construction of $A$ (a translate of $S\setminus\{a\}$) and verify $A\cap S=\emptyset$ and $(A+A)\cap S=\emptyset$ using the Sidon property.
  • Disproof track (general case): Try to build a sum-free $S$ of size $\mathfrak c$ so "additively thick" that for every $A\subseteq\mathbb R\setminus S$ of size $\mathfrak c$, one has $(A+A)\cap S\neq\emptyset$. No such construction is known to me from the given text, so I will end UNRESOLVED for the literal statement unless I find an explicit counterexample.

WORK

Fast reality check (toy examples).
  • $S=(1,2)$ is sum-free since $a,b>1\Rightarrow a+b>2$, hence $a+b\notin(1,2)$. Taking $A=[0,0.4)$ gives $A\cap S=\emptyset$ and $A+A=[0,0.8)\cap S=\emptyset$.
  • If $S$ is countable, then one expects plenty of room to choose $A$ of size $\mathfrak c$ avoiding $S$ and $S$-hits in $A+A$.

Lemma 949.1 (small $S$ case; no Sidon or sum-free hypothesis needed). Let $S\subset\mathbb R$ satisfy $|S|<\mathfrak c$. Then there exists $A\subseteq\mathbb R\setminus S$ with $|A|=\mathfrak c$ such that $(A+A)\cap S=\emptyset$.

Proof. Consider the collection \[ \mathcal F:=\{A\subseteq\mathbb R:\ A\cap S=\emptyset\ \text{and}\ (A+A)\cap S=\emptyset\}, \] ordered by inclusion. If $\{A_i\}_{i\in I}$ is a chain in $\mathcal F$, then $A:=\bigcup_{i\in I}A_i$ satisfies $A\cap S=\emptyset$ and \[ (A+A)\cap S\subseteq \bigcup_{i\in I}(A_i+A_i)\cap S=\emptyset \] because for any $x,y\in A$ there exists $j\in I$ with $x,y\in A_j$ (chain property), hence $x+y\in A_j+A_j$. Thus every chain has an upper bound, so by Zorn's lemma there exists a maximal $A\in\mathcal F$.

We claim $|A|=\mathfrak c$. By maximality, for every $x\in\mathbb R\setminus A$ at least one of the following holds:

  1. $x\in S$ (otherwise $A\cup\{x\}$ would still avoid $S$),
  2. $2x\in S$ (otherwise $2x\notin S$ so the new sum $x+x$ would not hit $S$),
  3. there exists $a\in A$ with $a+x\in S$ (otherwise all new sums $a+x$ would avoid $S$).
Equivalently, \begin{equation}\label{wsref-0-65-71-3a-6d-61-78-69-6d-61-6c-2d-63-6f-76-65-72} \mathbb R\subseteq A\ \cup\ \bigcup_{a\in A}(S-a)\ \cup\ S\ \cup\ (S/2), \end{equation} where $S-a:=\{s-a:s\in S\}$ and $S/2:=\{s/2:s\in S\}$.

Since $|S|<\mathfrak c$, also $|S\cup (S/2)|<\mathfrak c$. Hence $|(\mathbb R\setminus S)\cap (\mathbb R\setminus S/2)|=\mathfrak c$. From (eq:maximal-cover), \[ (\mathbb R\setminus S)\cap (\mathbb R\setminus S/2)\subseteq A\cup \bigcup_{a\in A}(S-a). \] Now each translate $S-a$ has cardinality $|S|$, and there are $|A|$ many translates. Therefore \[ \left|A\cup \bigcup_{a\in A}(S-a)\right|\le |A|+|A|\,|S|. \] If $|A|<\mathfrak c$, then (since $|S|<\mathfrak c$ and infinite-cardinal arithmetic gives $|A|+|A||S|=\max(|A|,|S|)<\mathfrak c$) the right-hand side would be $<\mathfrak c$, contradicting that the left-hand side has size $\ge \mathfrak c$. So $|A|=\mathfrak c$. $\square$

Lemma 949.2 (Sidon sets of size continuum; explicit construction). Assume $S\subset\mathbb R$ is Sidon and $|S|=\mathfrak c$. If $S$ contains some $a\ne 0$, define \[ A:=\left((S\setminus\{a\})-a/2\right)\setminus S. \] Then $|A|=\mathfrak c$, $A\cap S=\emptyset$, and $(A+A)\cap S=\emptyset$.

Proof. By definition $A\cap S=\emptyset$. We first show $|A|=\mathfrak c$. It suffices to show that \begin{equation}\label{wsref-0-65-71-3a-69-6e-74-65-72-73-65-63-74-69-6f-6e-2d-61-74-2d-6d-6f-73-74-2d-6f-6e-65} \left((S\setminus\{a\})-a/2\right)\cap S \ \text{has at most one element}. \end{equation} Indeed, the translate $(S\setminus\{a\})-a/2$ has cardinality $|S\setminus\{a\}|=\mathfrak c$, and removing a set of size at most one does not change cardinality $\mathfrak c$.

To prove (eq:intersection-at-most-one), suppose $x$ lies in the intersection. Then $x\in S$ and $x=s-a/2$ for some $s\in S\setminus\{a\}$, i.e. $s=x+a/2\in S$. Thus $x,\ x+a/2\in S$. If there were two distinct such $x_1\ne x_2$ in the intersection, we would have \[ (x_1+a/2)+x_2 = (x_2+a/2)+x_1, \] with all four terms in $S$. By the Sidon property, the unordered pairs must coincide: \[ \{x_1+a/2,\ x_2\}=\{x_2+a/2,\ x_1\}. \] Since $a\ne 0$, we cannot have $x_1+a/2=x_1$ or $x_2+a/2=x_2$. The only way the sets coincide is then $x_1+a/2=x_2$ and $x_2+a/2=x_1$, which implies $a=0$, contradiction. Hence the intersection has at most one element, proving (eq:intersection-at-most-one) and therefore $|A|=\mathfrak c$.

Next we prove $(A+A)\cap S=\emptyset$. Take any $x,y\in A$. Then there exist $s,t\in S\setminus\{a\}$ with \[ x=s-a/2,\qquad y=t-a/2. \] So \[ x+y = s+t-a. \] Suppose for contradiction that $x+y\in S$. Then we have an additive relation in $S$: \[ (s+t) = (x+y)+a, \] where $s,t\in S\setminus\{a\}$ and $(x+y),a\in S$. By the Sidon property applied to $s+t=(x+y)+a$, we must have \[ \{s,t\}=\{x+y,a\}. \] But $a\notin\{s,t\}$ by construction ($s,t\in S\setminus\{a\}$), contradiction. Therefore $x+y\notin S$ for all $x,y\in A$, i.e. $(A+A)\cap S=\emptyset$. $\square$

Corollary 949.3 (Sidon variant is true). If $S\subset\mathbb R$ is Sidon, then there exists $A\subseteq\mathbb R\setminus S$ with $|A|=\mathfrak c$ such that $A+A\subseteq\mathbb R\setminus S$.

Proof. If $|S|<\mathfrak c$, apply Lemma 949.1. If $|S|=\mathfrak c$ and $S$ contains some $a\ne0$, apply Lemma 949.2. (If $S=\{0\}$, take any $A$ of size $\mathfrak c$ avoiding $0$ and with $0\notin A+A$, e.g. $A\subset (0,\infty)$.) $\square$

VERIFICATION

  • Lemma 949.1: Checked that unions of chains preserve $(A+A)\cap S=\emptyset$ using the chain property (both summands land in some $A_j$). The maximality covering (eq:maximal-cover) was derived by contrapositive: if $x$ avoids $S$, $S/2$, and all $S-a$, then adding $x$ keeps sums out of $S$.
  • Cardinal arithmetic in Lemma 949.1 uses only that if $\kappa,\mu<\mathfrak c$ are infinite cardinals, then $\kappa+\kappa\mu=\max(\kappa,\mu)<\mathfrak c$; this is valid in ZFC.
  • Lemma 949.2: Verified the "intersection at most one" claim uses Sidon and $a\ne0$; if $a=0$ then construction degenerates. The $A+A$ disjointness uses that $a\notin S\setminus\{a\}$.
  • Corollary 949.3: handled the degenerate case $S=\{0\}$ separately.

FINAL

UNRESOLVED

(i) Strongest proved partial result. If $|S|<\mathfrak c$ (no further hypotheses), there exists $A\subseteq\mathbb R\setminus S$ with $|A|=\mathfrak c$ and $A+A\subseteq\mathbb R\setminus S$ (Lemma 949.1). Moreover, if $S$ is Sidon (any cardinality), the answer is yes (Corollary 949.3).

(ii) First gap. For a general sum-free set $S\subset\mathbb R$ of cardinality $\mathfrak c$ (no Sidon assumption), show either:

there exists $A\subseteq\mathbb R\setminus S$ with $|A|=\mathfrak c$ and $(A+A)\cap S=\emptyset$,
or construct an explicit counterexample $S$.

(iii) Top 3 next moves.

  1. Try to adapt Lemma 949.2 without Sidon: find a structured large subset $T\subseteq \mathbb R\setminus S$ with $T+T$ avoiding $S$ by choosing a translate/dilate $T=U-t$ for some large $U\subseteq S$.
  2. Attempt a counterexample: build sum-free $S$ of size $\mathfrak c$ such that for every continuum-sized $A\subseteq \mathbb R\setminus S$ one has $(A+A)\cap S\ne\emptyset$ ("$S$ hits every large sumset").
  3. Study maximal (under inclusion) sum-free sets $S$ in $(\mathbb R,+)$ and analyze the structure of $\mathbb R\setminus S$; determine whether maximality forces existence of the required $A$.

(iv) What a minimal counterexample would likely look like. A counterexample (if it exists) must have $|S|=\mathfrak c$ and must accumulate at $0$ (otherwise a sufficiently small neighborhood of $0$ would give an $A$ with $A+A$ still inside that neighborhood and disjoint from $S$). It must also be additively "thick" in the sense that $S$ intersects $A+A$ for every continuum-sized $A\subseteq\mathbb R\setminus S$.