Chains, Antichains, Sperner and Dilworth — Examples

Listing subsets by cardinality gives rank sizes $1,4,6,4,1$, and [L1] gives width $6$.

Deleting one element from a member of $\mathcal F$ gives exactly $\{1\},\{2\},\{3\}$, while adjoining one element gives exactly $123,124,134$. Thus the displayed shadows are correct.

The following symmetric chains partition all sixteen subsets: $\varnothing\subset1\subset12\subset123\subset1234$; $4\subset14\subset124$; $2\subset23\subset234$; $24$; $3\subset13\subset134$; and $34$. Their endpoint ranks sum to $4$, in agreement with [L2].

Steps 1.1, 1.2, and 1.3 verify the ranks, width, shadows, and an explicit symmetric chain decomposition.

The set $\{a_1,a_2,a_3\}$ is an antichain, so the width is at least $3$.

Every antichain contains at most one element from each comparable pair $\{a_i,b_i\}$, so it has at most $3$ elements. Hence the width is exactly $3$.

The three displayed two-element chains cover all six elements, so they form a chain cover of cardinality $3$.

By steps 1.2 and 1.3, and equivalently by [L1], the displayed cover has the minimum possible number of chains.

By [L1], the exponent-triple description is unique, and $d\mid e$ exactly when every exponent of $d$ is at most the corresponding exponent of $e$.

Componentwise minimum and maximum give the greatest common divisor and least common multiple, and the distributive identities hold coordinatewise for minimum and maximum on chains. Thus the divisor poset is a finite distributive lattice.

Its join-irreducibles are $2,4,3,5$. In their inherited order, $2<4$ and $3,5$ are incomparable with these and with each other.

A divisor $d$ maps to the order ideal of join-irreducibles dividing it: its $2$-exponent chooses $\varnothing$, $\{2\}$, or $\{2,4\}$, while its $3$- and $5$-exponents independently choose whether to include $3$ and $5$. This is exactly the Birkhoff map of [L2].

Hence the divisors of $60$ concretely realize Birkhoff's representation as the order ideals of the poset $2<4$ with isolated elements $3$ and $5$.

Every displayed set contains $\{1,2\}$, and outside this pair their elements lie in disjoint two-element blocks.

Therefore the intersection of any two distinct displayed sets is exactly $\{1,2\}$.

By [F1], the four sets form a sunflower with the stated core and petals.

Any two members of $\mathcal S_a$ intersect at $a$, so the star is intersecting.

The map $S\mapsto S\setminus\{a\}$ is a bijection from $\mathcal S_a$ to the $(k-1)$-subsets of $A\setminus\{a\}$. Hence $|\mathcal S_a|=\binom{n-1}{k-1}$.

By [L1], step 1.2 equals the universal upper bound, so the star is extremal.

The singleton $\{a\}$ is an antichain and cannot be enlarged, since both $b$ and $c$ are comparable with $a$.

The pair $\{b,c\}$ is an antichain. No three-element antichain exists, because the only three-element subset is $P$ itself and it contains the comparable pair $a,b$; hence the width of $P$ is $2$.

Thus $\{a\}$ is maximal of cardinality $1$ but not maximum, providing the required counterexample.

If $S,T\in[A]^k$ were disjoint, then $|S\cup T|=2k>n=|A|$, impossible. Hence the entire level $\mathcal F$ is intersecting.

By [L1], $|\mathcal F|=\binom nk=(n/k)\binom{n-1}{k-1}$. Since $k<n$, the factor $n/k$ is greater than $1$, so $|\mathcal F|>\binom{n-1}{k-1}$.

Thus for every $k<n<2k$, the full $k$th level is an intersecting family larger than a star, refuting the bound outside its stated range.

In $M_3$, one has $b\vee c=1$, $a\wedge b=0$, and $a\wedge c=0$. Therefore $a\wedge(b\vee c)=a$, while $(a\wedge b)\vee(a\wedge c)=0$.

In $N_5$, one has $a\vee c=1$, $b\wedge a=a$, and $b\wedge c=0$. Therefore $b\wedge(a\vee c)=b$, while $(b\wedge a)\vee(b\wedge c)=a$.

Since $a\ne0$ in $M_3$ and $a\ne b$ in $N_5$, each lattice violates the distributive identity in [F1]. Both are finite, so either one refutes the Statement.