gpt pro 5.2 attempt on Math-931

Problem 776: many sizes in an antichain with multiplicities

1) Formal restatement

Let $\mathcal{F}\subseteq 2^{[n]}$ be an antichain. Fix $r\in\mathbb{N}$ and assume that for every size $k$ occurring in $\mathcal{F}$, there are at least $r$ sets in $\mathcal{F}$ of size $k$. How large can $|\{|A|:A\in\mathcal{F}\}|$ be? For $r=1$ the maximum is $n-2$. For $r>1$ it is stated that $n-2$ is impossible but $n-3$ is achievable for large $n$. Question: determine/estimate how large $n$ must be (as a function of $r$) to guarantee an example with $n-3$ distinct sizes.

2) Quick literature/context check (no more than 8 lines)

• For $r=1$, there are explicit constructions with $n-2$ distinct sizes.
• For $r>1$, one loses at least one additional size (so $n-2$ is impossible).
• The quantitative threshold $n(r)$ for achieving $n-3$ sizes is the open part.

3) Attack plan

• Prove $n-2$ distinct sizes are impossible when $r>1$ by showing sizes $1$ and $n-1$ cannot coexist.
• Give an explicit construction achieving sizes $2,3,\dots,n-2$ (i.e. $n-3$ sizes) once $n$ is sufficiently large in terms of $r$; this yields an explicit upper bound $n(r)\le Cr+O(1)$.

4) Work

Impossibility of $n-2$ sizes for $r>1$.

In any antichain with more than one set, sizes $0$ and $n$ are absent (since $\varnothing\subseteq A\subseteq [n]$ for all $A$). Thus achieving $n-2$ sizes would force all but one of the sizes $1,2,\dots,n-1$ to occur, in particular both sizes $1$ and $n-1$.

If $r>1$, the presence of size $1$ means $\mathcal{F}$ contains two distinct singletons $\{a\}$ and $\{b\}$. Any $(n-1)$-set omits exactly one element, hence contains at least one of $a,b$. So any $(n-1)$-set $X\in\mathcal{F}$ would contain $\{a\}$ or $\{b\}$, contradicting antichain-ness. Therefore sizes $1$ and $n-1$ cannot both occur, and $n-2$ distinct sizes are impossible for $r>1$.

A linear-in-$r$ construction achieving $n-3$ sizes.

We construct an antichain $\mathcal{F}\subseteq 2^{[n]}$ with at least $r$ sets of each size $2,3,\dots,n-2$, provided $n\ge 8r+10$. Set $t=\lfloor n/2\rfloor$.

Step 1: build an intersecting antichain $\mathcal{G}$ with sizes $2,3,\dots,t$. Use a common element $1$. Let \[ A=\{2,3,\dots,r+1\},\qquad C=\{r+2,\dots,r+t-1\}, \] and let $B=\{r+t,\dots,n\}$ be the remaining elements. Then $|C|=t-2$ and $|B|=b=n-r-t+1$.

Define $r$ size-2 sets: \[ G_{2,i}=\{1,i+1\}\qquad(i=1,\dots,r). \] For each $s\in\{3,\dots,t\}$, let $F_s=\{r+2,\dots,r+s-2\}\subseteq C$ (so $|F_s|=s-3$). Assign to each pair $(s,i)$ (with $3\le s\le t$ and $1\le i\le r$) a distinct label $P_{s,i}\in\binom{B}{2}$, and set \[ G_{s,i}=\{1\}\cup P_{s,i}\cup F_s. \] Let $\mathcal{G}=\{G_{s,i}:2\le s\le t,\ 1\le i\le r\}$.

Step 2: $\mathcal{G}$ is an intersecting antichain. All sets contain $1$, so $\mathcal{G}$ is intersecting. No size-2 set $G_{2,i}$ is contained in any $G_{s,i'}$ with $s\ge 3$, because $i+1\in A$ and all $G_{s,i'}$ (for $s\ge 3$) use only $\{1\}\cup B\cup C$, avoiding $A\setminus\{1\}$. If $s,s'\ge 3$ and $G_{s,i}\subseteq G_{s',i'}$, then their intersections with $B$ satisfy $P_{s,i}\subseteq P_{s',i'}$, forcing $P_{s,i}=P_{s',i'}$, contradicting distinctness. Hence $\mathcal{G}$ is an antichain.

Step 3: add complements. Let \[ \mathcal{F}=\mathcal{G}\ \cup\ \{[n]\setminus G: G\in\mathcal{G}\}. \] Because $\mathcal{G}$ is intersecting, no $G\in\mathcal{G}$ is contained in the complement of another member, so $\mathcal{F}$ is still an antichain. The sizes appearing in $\mathcal{F}$ are exactly $2,3,\dots,n-2$, i.e. $n-3$ distinct sizes, and each such size occurs at least $r$ times (from $\mathcal{G}$ or from complements).

Step 4: enough labels. We need $\binom{b}{2}\ge r(t-2)$ to assign distinct pairs $P_{s,i}$. For $n\ge 8r+10$ one checks that $b=n-r-t+1\ge 3r+6$ and $\binom{b}{2}$ dominates $r(t-2)$, so the labeling is feasible. Thus $n(r)\le 8r+10$ is an explicit sufficient bound.

5) Verification

• The $n-2$ impossibility uses only $r>1$: two singletons force containment into any $(n-1)$-set.
• The construction is checked by separating (i) size-2 sets using reserved elements, and (ii) larger sets distinguished by unique 2-element labels in $B$.
• Intersecting $\mathcal{G}$ guarantees no cross-containment when adding complements.

6) Final (exactly one label and one sub-label)

LABEL: UNRESOLVED.
SUBLABEL: Proved $n-2$ sizes impossible for $r>1$; gave an explicit construction showing $n(r)\le 8r+10$ suffices for $n-3$ sizes.

7) Completion estimate

Determining the true threshold $n(r)$ would require either improved constructions (reducing the constant $8$) or nontrivial lower bounds showing $n$ must grow at least linearly (or faster) in $r$.