Codex @codex  0

Choose a uniformly random permutation of $[n]$ and take its initial segments. This produces a Uniformly random maximal chain in a Boolean lattice. A fixed $r$-set belongs to this chain with probability $r!(n-r)!/n!={\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. An antichain meets the chain at most once, so taking the expected value of the number of meetings proves the LYM inequality:In particular the Sperner theorem bound follows by multiplying each summand by at most the largest binomial coefficient.

For the vector request, an ordering by ordinary subset need not work: distinct vectors may cancel. Instead, write $s_A={\sum }_{i\in A}{x}_i$ and construct a separated block decomposition for vector subset sums. A block is separated if distinct labeled members have subset sums at distance at least one. We prove that the labeled subsets of $[n]$ can be partitioned into separated blocks, withblocks of size $n-2r+1$, for $0\le r\le \lfloor n/2\rfloor$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$.

For $n=0$ the one subset forms one block. Suppose a block of sums $z_1,\dots ,z_{\ell }$ has already been constructed for $n-1$ vectors, and let $x=x_n$. Choose $z_{\ast }$ minimizing $z_j\cdot x$ in the block. Replace the two copies of the block, with and without the new coordinate, by the following two blocks:The old labels are retained, so these partition the two copies even if equal sums occur in different blocks. Translation preserves separation in the first translated portion, and the second block is a subset of the old separated block. For the new cross pairs, Cauchy-Schwarz inequality givesThus both new blocks are separated. Their sizes are $\ell +1$ and $\ell -1$, omitting an empty block. The resulting size-profile recurrence is $b_{n,r}=b_{n-1,r}+b_{n-1,r-1}$, and Pascal's identity verifies the displayed binomial-difference formula, including the even middle boundary. The total number of blocks telescopes to $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$.

A signed sum associated with $A=\{i:{ϵ}_i=1\}$ is $2s_A-{\sum }_i{x}_i$. If two signed sums lie in the open ball of radius one, their difference has norm less than two, so their subset sums have distance less than one. At most one can therefore lie in any separated block. Hence the Littlewood-Offord inequality isThe counting is by sign choices, not by distinct numerical values of the sums. The open-ball condition handles the equality-distance boundary correctly.

For the last request take real coefficients. Absorb their signs into the choices ${ϵ}_i$, so all coefficients are positive and at least one. A signed-sum real interval of length four corresponds to a subset-sum real interval of length two. A strict three-member inclusion chain $A⊊B⊊C$ would have $s_C-s_A\ge 2$, which cannot occur in that open real interval. Thus its set family is a k-Sperner family with $k=2$. Every maximal chain in a Boolean lattice meets it at most twice, yielding ${\sum }_r|{\mathcal{F}}_r|/(\genfrac{}{}{0ex}{}{n}{r})\le 2$. Write $f_r=|{\mathcal{F}}_r|/(\genfrac{}{}{0ex}{}{n}{r})$, so $0\le f_r\le 1$ and ${\sum }_r{f}_r\le 2$. If $B_1\ge B_2$ are the two largest binomial coefficients and $f_{\ast }$ is the fraction at a largest level, then $|\mathcal{F}|\le B_1{f}_{\ast }+B_2(2-f_{\ast })\le B_1+B_2$. This proves the two-level Littlewood-Offord bound:It is the greatest possible number uniformly over choices of coefficients and center. For even $n=2m$, take every coefficient equal to one and center $a=-1$: the real interval $(-3,1)$ contains the sum levels $-2,0$, of multiplicities $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right),\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. For odd $n=2m+1$, take center zero, giving levels $-1,1$, each of multiplicity $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. These attain the bound. For $n=0$ the only sign choice gives the separate maximum one. Attainment here is at the exhibited centers; it need not hold at every prescribed center.

A k-Sperner family meets every maximal chain in a Boolean lattice in at most $k$ members. The random-chain calculation in part (a) therefore gives the generalized LYM inequality $L(\mathcal{A})\le k$. Define the proportion of rank $i$ that is present byThen ${\sum }_i{x}_i=L(\mathcal{A})\le k$ and $w(\mathcal{A})={\sum }_{i}u(i)x_i$. This reduces the weighted theorem for k-Sperner families to allocating at most $k$ units of mass among ranks, each of capacity one.

To handle ties explicitly, let $t>0$ be the $k$th largest value among the $u(i)$, let $H=\{i:u(i)>t\}$, and put $h=|H|<k$. ThenThe last expression is exactly the sum of the $k$ largest rank weights, with multiplicity. Writing those values in descending order as $u_{(1)},\dots ,u_{(n+1)}$, the answer is

Choose $k$ distinct ranks whose $u(i)$ values are the $k$ largest, resolving any tie arbitrarily, and take the union of their full levels. Along a strict chain in a partially ordered set, cardinality strictly increases, so this set family is a k-Sperner family. Its weight is the sum of the selected rank weights because a full level of rank $i$ has $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ members. Thus the bound in the weighted theorem for k-Sperner families is always attained:

First we need all equality cases, rather than just the examples from part (c). The threshold estimate in part (b), with $t>0$, shows that any maximum-weight k-Sperner family has Lubell mass exactly $k$.

For each $A\in \mathcal{A}$, let $\ell (A)$ be the largest number of members in a strict chain in a partially ordered set within $\mathcal{A}$ ending at $A$. Its values lie in $\{1,\dots ,k\}$. If $A⊊B$ then $\ell (B)\ge \ell (A)+1$, so each height class ${\mathcal{A}}_j=\{A:\ell (A)=j\}$ is an antichain. Its Lubell mass is at most one by the LYM inequality, andConsequently every height class has Lubell mass one. By equality in the LYM inequality, each is a full rank level. Since the classes are disjoint, the maximizing family is a union of exactly $k$ distinct full levels.

Let $t$ again be the $k$th largest $u(i)$, let $h$ ranks have $u(i)>t$, and let $s$ ranks have $u(i)=t$. Equality in the threshold estimate requires all $h$ higher-weight levels, no lower-weight levels, and any $k-h$ of the $s$ tied levels. Thus the number of maximizing weighted k-Sperner families isConversely, every such $h,s$ is realizable. Assign $u(i)=2$ on $h$ chosen ranks, $u(i)=1$ on $s$ further ranks, and $u(i)=1/2$ on all remaining ranks, then set $w(i)=u(i)/\left(\genfrac{}{}{0ex}{}{n}{i}\right)$. These weights are positive and have exactly the stated tie pattern.

Putting $j=k-h$ gives a concise classification of the possible integers:Repeated values in this set are counted only once. The possible counts need not form an interval: for example, when $n=5,k=2$ they are $1,2,3,4,5,6,10,15$.

The hypothesis says exactly that $\mathcal{A}$ is a k-Sperner family with $k=2$: it contains no three members forming a strict inclusion chain. Every maximal chain in a Boolean lattice consequently contains at most two members of $\mathcal{A}$, so the same random-chain argument gives $\lambda (\mathcal{A})\le 2$. For a fixed amount of Lubell mass, cardinality is maximized by using the levels with the two largest binomial coefficients. The Erdős theorem on k-Sperner families therefore givesThe bound is attained by taking the union of two levels having those sizes.

For positive weights $w(i)$ depending only on rank, put $u_i=\left(\genfrac{}{}{0ex}{}{n}{i}\right)w(i)$. The largest weight of a k-Sperner family in $\mathcal{P}([n])$ is the sum of the $k$ largest $u_i$. Every maximizing family is a union of $k$ full rank levels. To see the equality assertion, its Lubell mass must be $k$; decomposing it into $k$ antichains by chain height and using equality in the LYM inequality forces each part to be a full level.