Erdős Problem #447
Status Solved · erdosproblems.comLLM status Solved from source status
Erdős problems openaigpt 6 astra ultra 65% cataloguedSolvedSolvedStatement
Statement and scope.
A family $\mathcal F\subseteq2^{[n]}$ is union-free if no three distinct members satisfy $A\cup B=C$. Must $|\mathcal F|=o(2^n)$, and even $|\mathcal F|\leq(1+o(1))\binom n{\lfloor n/2\rfloor}$? We prove the first assertion and the correct order of magnitude, with upper constant $2\sqrt2$ in place of $1$. The sharp leading constant is unresolved in this attempt.Symmetric chains, with their construction.
The Boolean lattice $2^{[m]}$ can be partitioned into inclusion chains whose endpoint ranks add to $m$. Induct on $m$. For a chain $x_k\subset x_{k+1}\subset\cdots\subset x_{m-k}$, adjoining a new coordinate partitions its two copies into \[ x_k,\ldots,x_{m-k},x_{m-k}\cup\{m+1\} \] and, when nonempty, \[ x_k\cup\{m+1\},\ldots,x_{m-k-1}\cup\{m+1\}. \] The endpoint ranks in each new chain add to $m+1$. This establishes the induction from the one-element lattice at $m=0$. Each symmetric chain has exactly one member of rank $\lfloor m/2\rfloor$, so their number is \[ c_m=\binom m{\lfloor m/2\rfloor}. \tag{1} \]A union-free family on a product of two chains.
Split the ground set into disjoint parts of sizes $m=\lfloor n/2\rfloor$ and $\ell=\lceil n/2\rceil$, and use the chain partitions above on each part. The product of a chain of length $r$ and one of length $s$ is an $r$-by-$s$ grid, with union given by coordinatewise maximum. If an occupied cell of $\mathcal F$ is not the first occupied cell in its row or its column, there are occupied cells earlier in its row and earlier in its column. These are distinct from one another and their union is the given cell, a contradiction. Every occupied cell must therefore be first in its row or first in its column. There are at most $r+s$ such cells.Summing over all products of chains gives the exact bound \[ |\mathcal F|\leq c_\ell 2^m+c_m2^\ell. \tag{2} \] Here the sum of chain lengths in $2^{[m]}$ is $2^m$, and there are $c_\ell$ choices of the other chain, and conversely. Stirling's formula applied to $1$ yields \[ |\mathcal F|\leq (2\sqrt2+o(1))\binom n{\lfloor n/2\rfloor} =O(2^n/\sqrt n)=o(2^n). \tag{3} \]
A matching lower order and the remaining gap.
All subsets of cardinality $\lfloor n/2\rfloor$ form a union-free family: the union of two distinct such sets is strictly larger. Thus the maximum size is $\Theta(\binom n{\lfloor n/2\rfloor})$. The missing step is the improvement of the upper constant in $3$ from $2\sqrt2$ to $1$.Kleitman's known sharp theorem and the earlier constant $2\sqrt2$ are recorded in Balogh and Wagner, On the number of union-free families, Introduction and Theorem 1.1, https://arxiv.org/abs/1601.03659. Current statement: https://www.erdosproblems.com/447, checked 24 September 2026. The chain proof is supplied in full here; the sharper theorem is not assumed to fill the missing step.
Completion estimate.
The first question and the correct order are proved; the sharp leading constant is missing.Restatement from Math-5656-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-5656
Erdős Problem 447
Statement and scope.
A family $\mathcal F\subseteq2^{[n]}$ is union-free if no three distinct members satisfy $A\cup B=C$. Must $|\mathcal F|=o(2^n)$, and even $|\mathcal F|\leq(1+o(1))\binom n{\lfloor n/2\rfloor}$? We prove the first assertion and the correct order of magnitude, with upper constant $2\sqrt2$ in place of $1$. The sharp leading constant is unresolved in this attempt.Symmetric chains, with their construction.
The Boolean lattice $2^{[m]}$ can be partitioned into inclusion chains whose endpoint ranks add to $m$. Induct on $m$. For a chain $x_k\subset x_{k+1}\subset\cdots\subset x_{m-k}$, adjoining a new coordinate partitions its two copies into \[ x_k,\ldots,x_{m-k},x_{m-k}\cup\{m+1\} \] and, when nonempty, \[ x_k\cup\{m+1\},\ldots,x_{m-k-1}\cup\{m+1\}. \] The endpoint ranks in each new chain add to $m+1$. This establishes the induction from the one-element lattice at $m=0$. Each symmetric chain has exactly one member of rank $\lfloor m/2\rfloor$, so their number is \[ c_m=\binom m{\lfloor m/2\rfloor}. \tag{1} \]A union-free family on a product of two chains.
Split the ground set into disjoint parts of sizes $m=\lfloor n/2\rfloor$ and $\ell=\lceil n/2\rceil$, and use the chain partitions above on each part. The product of a chain of length $r$ and one of length $s$ is an $r$-by-$s$ grid, with union given by coordinatewise maximum. If an occupied cell of $\mathcal F$ is not the first occupied cell in its row or its column, there are occupied cells earlier in its row and earlier in its column. These are distinct from one another and their union is the given cell, a contradiction. Every occupied cell must therefore be first in its row or first in its column. There are at most $r+s$ such cells.Summing over all products of chains gives the exact bound \[ |\mathcal F|\leq c_\ell 2^m+c_m2^\ell. \tag{2} \] Here the sum of chain lengths in $2^{[m]}$ is $2^m$, and there are $c_\ell$ choices of the other chain, and conversely. Stirling's formula applied to $1$ yields \[ |\mathcal F|\leq (2\sqrt2+o(1))\binom n{\lfloor n/2\rfloor} =O(2^n/\sqrt n)=o(2^n). \tag{3} \]
A matching lower order and the remaining gap.
All subsets of cardinality $\lfloor n/2\rfloor$ form a union-free family: the union of two distinct such sets is strictly larger. Thus the maximum size is $\Theta(\binom n{\lfloor n/2\rfloor})$. The missing step is the improvement of the upper constant in $3$ from $2\sqrt2$ to $1$.Kleitman's known sharp theorem and the earlier constant $2\sqrt2$ are recorded in Balogh and Wagner, On the number of union-free families, Introduction and Theorem 1.1, https://arxiv.org/abs/1601.03659. Current statement: https://www.erdosproblems.com/447, checked 24 September 2026. The chain proof is supplied in full here; the sharper theorem is not assumed to fill the missing step.