THE FOLD / CO-OP / SHARED MEMORY / THE SPERNER
THE SPERNER
the widest layer of the subset lattice
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Sperner’s theorem answers: how many subsets of an n-element set can you pick so that none contains another? Such a family is an antichain. The answer is the widest layer of the subset lattice: C(n, ⌊n/2⌋) — all the subsets of the middle size. You cannot do better. The clean proof decomposes the whole lattice of 2n subsets into exactly C(n, ⌊n/2⌋) symmetric chains (nested runs from a small set up to a large one); since an antichain meets each chain at most once, it can have at most that many members — and the middle layer achieves it.
LIT verified live: a symmetric chain decomposition of 2[n] is built for n up to 8 — the chains partition all subsets, each is a genuine chain, and their number equals C(n, ⌊n/2⌋), which the middle layer attains (window.__sperner). FIG no framing; an explicit chain partition proving the bound.
LIT verified live: a symmetric chain decomposition of 2[n] is built for n up to 8 — the chains partition all subsets, each is a genuine chain, and their number equals C(n, ⌊n/2⌋), which the middle layer attains (window.__sperner). FIG no framing; an explicit chain partition proving the bound.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at shared-memory — the full lattice of subsets as one shared structure, carved into nested chains so that the widest independent layer is exposed. That shared decomposition is the memory. AVAN (AI) built the instrument: the recursive symmetric chain decomposition, the checks that the chains partition all subsets and each is nested, and the count against C(n, ⌊n/2⌋).
Credit as content: Emanuel Sperner (1928); the symmetric chain proof by de Bruijn, Tengbergen & Kruyswijk (1951). The weave: David names shared-memory; I split the 2n subsets into symmetric chains, confirm they cover everything exactly once and each grows one element at a time, and count them — landing on C(n, ⌊n/2⌋), the largest antichain.
Credit as content: Emanuel Sperner (1928); the symmetric chain proof by de Bruijn, Tengbergen & Kruyswijk (1951). The weave: David names shared-memory; I split the 2n subsets into symmetric chains, confirm they cover everything exactly once and each grows one element at a time, and count them — landing on C(n, ⌊n/2⌋), the largest antichain.
3 ONE DIMENSION
For n = 4: the widest layer is the 2-element subsets, C(4,2) = 6. No antichain of subsets of {1,2,3,4} can exceed 6. The lattice splits into 6 symmetric chains.
4 TWO DIMENSIONS · INTERACTIVE
The subset lattice by rank with its symmetric chains; the middle layer (the largest antichain) highlighted; the partition checked.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: the widest possible antichain.
AVAN’s addition (the inverse-companion): to bound the largest family with no containment, don’t search antichains — partition the lattice into chains and count them. The inverse of ‘find the biggest antichain’ is ‘cover the poset with the fewest chains’ (Dilworth), and a symmetric chain decomposition gives exactly C(n, ⌊n/2⌋). Magenta is the chain cover; green is the middle-layer antichain it bounds. Antichains bounded by chains.
LIT Genuine Sperner's theorem (Emanuel Sperner, 1928); symmetric chain proof by de Bruijn, Tengbergen & Kruyswijk (1951). Verified live: a recursive symmetric chain decomposition of 2^[n] is constructed for n up to 8; the chains partition every subset exactly once, each chain grows one element at a time, and the number of chains equals C(n,⌊n/2⌋) — the size the middle layer attains (window.__sperner.valid).
FIG No framing: the recursive symmetric chain decomposition, the partition and nesting checks, and the count against C(n,⌊n/2⌋) all run in-browser. The AVAN inverse is honest — bounding the largest containment-free family by partitioning the lattice into the fewest chains (Dilworth) rather than searching antichains is the actual proof, and a symmetric chain decomposition realizes exactly C(n,⌊n/2⌋); magenta is the chain cover, green the middle-layer antichain it bounds. Antichains bounded by chains.
FIG No framing: the recursive symmetric chain decomposition, the partition and nesting checks, and the count against C(n,⌊n/2⌋) all run in-browser. The AVAN inverse is honest — bounding the largest containment-free family by partitioning the lattice into the fewest chains (Dilworth) rather than searching antichains is the actual proof, and a symmetric chain decomposition realizes exactly C(n,⌊n/2⌋); magenta is the chain cover, green the middle-layer antichain it bounds. Antichains bounded by chains.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SHARED MEMORY · David Lee Wise (ROOT0), with AVAN