Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. The answer is yes, with : the largest family of subsets of in which no member is the union of other members satisfies
The argument has two lines. The middle layer, all subsets of size , has no member that is a union of other members, because a union of two or more distinct sets of that size is strictly larger, so . In the other direction, a family with no member a union of other members has in particular no solution of in distinct members, so it is union-free in the sense of Problem 447, and Kleitman's theorem [Kl71], the solution of that problem, gives . The two bounds match, and Stirling's formula turns the central binomial coefficient into . The printed source of the problem [Er71, item 18] gives the unpublished Erdős–Kleitman bounds and the conjectured asymptotic with the exponent in place of ; the middle layer shows that cannot be right, so the conjecture is false as printed, and the site reads the exponent as a misprint. The claim is the corrected form of the printed conjecture.
Submission note. Posted to the site's forum by Zach Hunter on 13 September 2025:
actually this solves the problem. by problem 447, any such family must have size at most . and we still have this as a lower bound.
(The site has been updated to address this comment.)
Depends on. Kleitman 1971, the accepted claim of Problem 447, supplies the upper bound: Kleitman's theorem that a union-free family of subsets of an -set has at most members.
Claimant. Zach Hunter, posting in the problem's discussion thread on erdosproblems.com under the username zach hunter on 13 September 2025: a first comment gives the middle-layer lower bound and a second, minutes later, the deduction of the upper bound from Problem 447 and the conclusion. The result is a thread observation and not a manuscript; it has a page because it is the result the site credits with the solution.
Acceptance. Reviewed: Thomas Bloom, the site's curator, replied in the thread the same day, thanking both posters and adopting the reading of the exponent in [Er71] as a misprint; the site's problem page labels the problem solved and credits Hunter's observation that the solution of Problem 447 implies , which is where the deduction itself is accepted. No refereed publication of the deduction exists; it is two lines over Kleitman's refereed theorem.
Formalization. The Lean file among the links, in Boris Alexeev's repository
lean-proofs, declares itself a formalization of a solution to Problem 1023
whose original proof "was found by: Kleitman and Hunter", and says that
Kleitman's proof was auto-formalized by Aristotle (Harmonic). It imports the
repository's development for Problem 447 (ErdosProblems.Erdos447, Kleitman's
theorem as auto-formalized by Aristotle) and uses its erdos_447, the
asymptotic for families with no
; a second header block describing the contents of that imported
development is carried over into this file. On top of the import it defines a
family to be union-free when no member is the join of a nonempty set of other
members, the problem's condition, defines as the largest size of such a
family of subsets of an -element type, carries Kleitman's asymptotic over
to the problem's families in kleitman_bound_many, identifies the constant
as in many_sqrt_two_div_pi, and ends with
theorem erdos_1023, the existence of with ,
followed by #print axioms whose recorded output lists propext,
Classical.choice and Quot.sound. Alexeev announced
the file in the thread on 10 February 2026; the link is pinned to the
repository's commit of that day. This corpus has not built the file, so the
formalization is a link and not formalized evidence. The formal-conjectures
statement file for the problem, as of its commit of 18 September 2026, marks
it solved and points at the same file of the same repository under a later
toolchain folder (v4.29.1), unpinned; a statement file is not a
formalization.