Combinatorics (Fall 2010)/Finite set systems

Systems of Distinct Representatives (SDR)

A system of distinct representatives (SDR) (also called a transversal) for a sequence of (not necessarily distinct) sets [math]\displaystyle{ S_1,S_2,\ldots,S_m }[/math] is a sequence of distinct elements [math]\displaystyle{ x_1,x_2,\ldots,x_m }[/math] such that [math]\displaystyle{ x_i\in S_i }[/math] for all [math]\displaystyle{ i=1,2,\ldots,m }[/math].

Hall's marriage theorem

If the sets [math]\displaystyle{ S_1,S_2,\ldots,S_m }[/math] have a system of distinct representatives [math]\displaystyle{ x_1\in S_1,x_2\in S_2,\ldots,x_m\in S_m }[/math], then it is obvious that [math]\displaystyle{ \left|S_1\cup S_2\cup\cdots\cup S_m\right|\ge |\{x_1,x_2,\ldots,x_m\}|=m }[/math]. Moreover, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,m\} }[/math],

[math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\ge |\{x_i\mid i\in I\}|=|I| }[/math]

because [math]\displaystyle{ x_1,x_2,\ldots,x_m }[/math] are distinct.

Hall's Theorem (SDR)

The sets [math]\displaystyle{ S_1,S_2,\ldots,S_m }[/math] have a system of distinct representatives (SDR) if and only if [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\ge |I| }[/math] for all [math]\displaystyle{ I\subseteq\{1,2,\ldots,m\} }[/math].

The condition that [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\ge |I| }[/math] for all [math]\displaystyle{ I\subseteq\{1,2,\ldots,m\} }[/math] is also called the Hall's condition.

Proof. The necessity of Hall's condition for the existence of SDR is obvious. Suppose that [math]\displaystyle{ x_1\in S_1, x_2\in S_2,\ldots,x_m\in S_m }[/math] is an SDR. [math]\displaystyle{ \forall I\subseteq\{1,2,\ldots,m\}, \bigcup_{i\in I}S_i\supseteq \{x_i\mid i\in I\} }[/math] thus [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\ge | \{x_i\mid i\in I\}|=|I| }[/math]. Now we prove the sufficiency of Hall's condition by induction on [math]\displaystyle{ m }[/math]. When [math]\displaystyle{ m=1 }[/math], the theorem trivially holds. Now assume the theorem hold for any integer smaller than [math]\displaystyle{ m }[/math]. We show the theorem hold for [math]\displaystyle{ m }[/math]. A subcollection of sets [math]\displaystyle{ \{S_i\mid i\in I\} }[/math], [math]\displaystyle{ |I|\lt m\, }[/math], is called a critical family if [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|=|I| }[/math]. Case.1: There is no critical family, i.e. for each [math]\displaystyle{ I\subset\{1,2,\ldots,m\} }[/math], [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\gt |I| }[/math]. Take [math]\displaystyle{ S_m }[/math] and choose an arbitrary [math]\displaystyle{ x\in S_m }[/math] as its representative. Remove [math]\displaystyle{ x }[/math] from all other sets by letting [math]\displaystyle{ S'_i=S_i\setminus \{x\} }[/math] for [math]\displaystyle{ 1\le i\le m-1 }[/math]. Then for all [math]\displaystyle{ I\subseteq\{1,2,\ldots,m-1\} }[/math], [math]\displaystyle{ \left|\bigcup_{i\in I}S_i'\right|\ge \left|\bigcup_{i\in I}S_i\right|-1\ge |I| }[/math]. Due to the induction hypothesis, [math]\displaystyle{ S_1,\ldots,S_m }[/math] have an SDR, say [math]\displaystyle{ x_1\in S_1,\ldots,x_{m-1}\in S_{m-1} }[/math]. It is obvious that none of them equals [math]\displaystyle{ x }[/math] because [math]\displaystyle{ x }[/math] is removed. Thus, [math]\displaystyle{ x_1,\ldots,x_{m-1} }[/math] and [math]\displaystyle{ x\, }[/math] form an SDR for [math]\displaystyle{ S_1,S_2,\ldots,S_m }[/math]. Case.1: There is a critical family, i.e. [math]\displaystyle{ \exists I\subset\{1,2,\ldots,m\}, |I|\lt m }[/math], such that [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|=|I| }[/math]. Suppose [math]\displaystyle{ S_{m-k+1},\ldots, S_m }[/math] are such a collection of [math]\displaystyle{ k\lt m }[/math] sets. Hall's condition certainly holds for these sets. Since [math]\displaystyle{ k\lt m }[/math], due to the induction hypothesis, there is an SDR for the sets, say [math]\displaystyle{ x_{m-k+1}\in S_{m-k+1},\ldots,x_m\in S_m }[/math]. Again, remove the [math]\displaystyle{ k }[/math] elements from the remaining sets by letting [math]\displaystyle{ S'_i=S_i\setminus\{x_{m-k+1},\ldots,x_m\} }[/math] for [math]\displaystyle{ 1\le i\le m-k }[/math]. By the Hall's condition, for any [math]\displaystyle{ I\subseteq\{1,2,\ldots,m-k\} }[/math], writing that [math]\displaystyle{ S=\bigcup_{i\in I}S_i\cup\bigcup_{i=m-k+1}^m S_i }[/math], [math]\displaystyle{ \left|S\right|\ge |I|+k }[/math], thus [math]\displaystyle{ \left|\bigcup_{i\in I}S_i'\right|\ge |S|-\left|\bigcup_{i=m-k+1}^m S_i\right|\ge |I| }[/math]. Due to the induction hypothesis, there is an SDR for [math]\displaystyle{ S_1,\ldots, S_{m-k} }[/math], say [math]\displaystyle{ x_1\in S_1,\ldots, x_{m-k}\in S_{m-k} }[/math]. Combining it with the SDR [math]\displaystyle{ x_{m-k+1},\ldots, x_{m} }[/math] for [math]\displaystyle{ S_{m-k+1},\ldots, S_{m} }[/math], we have an SDR for [math]\displaystyle{ S_{1},\ldots, S_{m} }[/math]. [math]\displaystyle{ \square }[/math]

Hall's Theorem (matching in bipartite graph)

A bipartite graph [math]\displaystyle{ G(U,V,E) }[/math] has a matching of [math]\displaystyle{ U }[/math] if and only if [math]\displaystyle{ \left|N(S)\right|\ge |S| }[/math] for all [math]\displaystyle{ S\subseteq U }[/math].

Doubly stochastic matrices

Theorem (Birkhoff 1949; von Neumann 1953)

Every doubly stochastic matrix is a convex combination of permutation matrices.

Proof. We prove a more general result: Every [math]\displaystyle{ n\times n }[/math] nonnegative matrix [math]\displaystyle{ A=(a_{ij}) }[/math] with [math]\displaystyle{ \forall i, \sum_{j=1}^na_{ij}=\gamma }[/math] and [math]\displaystyle{ \forall j, \sum_{i=1}^na_{ij}=\gamma }[/math], for some [math]\displaystyle{ \gamma\gt 0 }[/math], can be expressed as a linear combination [math]\displaystyle{ A=\sum_{i=1}^s\lambda_iP_i }[/math] of permutation matrices [math]\displaystyle{ P_1,\ldots,P_s }[/math], where [math]\displaystyle{ \lambda_1,\ldots,\lambda_{s} }[/math] are nonnegative reals such that [math]\displaystyle{ \sum_{i=1}^s\lambda_i=\gamma }[/math]. The Birkhoff-von Neumann theorem is a special case of above statement with [math]\displaystyle{ \gamma=1 }[/math]. We prove by induction on the number of non-zero entries in [math]\displaystyle{ A }[/math], denoted by [math]\displaystyle{ m }[/math]. Since all the row and column sums equal to some [math]\displaystyle{ \gamma\gt 0 }[/math], there are at least [math]\displaystyle{ n }[/math] such entries. If there are exactly [math]\displaystyle{ n }[/math] non-zero entries, the only possible [math]\displaystyle{ A }[/math] with [math]\displaystyle{ \forall i, \sum_{j=1}^na_{ij}=\gamma }[/math] and [math]\displaystyle{ \forall j, \sum_{i=1}^na_{ij}=\gamma }[/math] is that [math]\displaystyle{ A=\gamma P }[/math] for some permutation matrix [math]\displaystyle{ P }[/math]. Now suppose [math]\displaystyle{ A }[/math] has [math]\displaystyle{ m\gt n }[/math] non-zero entries and the theorem holds for matrices with a smaller number of non-zero entries. Define [math]\displaystyle{ S_i=\{j\mid a_{ij}\gt 0\},\quad i=1,2,\ldots,n, }[/math] and observe that [math]\displaystyle{ S_1, S_2,\ldots, S_n }[/math] satisfy Hall's condition. Otherwise if there exists an [math]\displaystyle{ I\subseteq\{1,2,\ldots,m\} }[/math] such that [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\lt |I| }[/math], then all the non-zero entries of the rows in [math]\displaystyle{ I }[/math] would occupy less than [math]\displaystyle{ |I| }[/math] columns; hence the sum of these entries [math]\displaystyle{ \sum_{i\in I}\sum_{j\in S_j}a_{ij}\le\sum_{j\in\bigcup_{i\in I}S_i}\sum_{i=1}^na_{ij}\lt |I|\gamma }[/math], which contradicts that [math]\displaystyle{ \sum_{i\in I}\sum_{j\in S_j}a_{ij}=\sum_{i\in I}\sum_{j=1}^na_{ij}=|I|\gamma }[/math]. By Hall's theorem, there is an SDR [math]\displaystyle{ j_1\in S_1,\ldots,j_n\in S_n }[/math]. Take the permutation matrix [math]\displaystyle{ P=(p_{ij}) }[/math] with entries [math]\displaystyle{ p_{ij}=1 }[/math] if and only if [math]\displaystyle{ j=j_i }[/math]. Let [math]\displaystyle{ \lambda=\min\{a_{ij}\mid j=j_i, i=1,2,\ldots,n\} }[/math], and consider the matrix [math]\displaystyle{ A'=A-\lambda P }[/math]. It is obvious that [math]\displaystyle{ A' }[/math] has less non-zero entries than [math]\displaystyle{ A }[/math] has. Moreover, [math]\displaystyle{ A' }[/math] satisfies that [math]\displaystyle{ \forall i, \sum_{j=1}^na'_{ij}=\gamma-\lambda }[/math] and [math]\displaystyle{ \forall j, \sum_{i=1}^na'_{ij}=\gamma-\lambda }[/math]. Apply the induction hypothesis, [math]\displaystyle{ A'=\sum_{i=1}^s\lambda_iP_i }[/math] for permutation matrices [math]\displaystyle{ P_1,\ldots,P_m }[/math] and [math]\displaystyle{ \lambda_1,\ldots,\lambda_s\gt 0 }[/math] with [math]\displaystyle{ \sum_{i=1}^s\lambda_i=\gamma-\lambda }[/math]. Therefore, [math]\displaystyle{ A=\lambda P+A'=\lambda P+\sum_{i=1}^s\lambda_iP_i }[/math], where [math]\displaystyle{ \lambda+\sum_{i=1}^s\lambda_i=\lambda+(\gamma-\lambda)=\gamma }[/math]. [math]\displaystyle{ \square }[/math]

Min-max theorems

• König-Egerváry theorem (König 1931; Egerváry 1931): in a bipartite graph, the maximum number of edges in a matching equals the minimum number of vertices in a vertex cover.
• Menger's theorem (Menger 1927): the minimum number of vertices separating two given vertices in a graph equals the maximum number of vertex-disjoint paths between the two vertices.
• Dilworth's theorem (Dilworth 1950): the minimum number of chains which cover a partially ordered set equals the maximum number of elements in an antichain.

König-Egerváry Theorem (graph theory form)

In any bipartite graph, the size of a maximum matching equals the size of a minimum vertex cover.

König-Egerváry Theorem (matrix form)

Let [math]\displaystyle{ A }[/math] be an [math]\displaystyle{ m\times n }[/math] 0-1 matrix. The maximum number of independent 1's is equal to the minimum number of rows and columns required to cover all the 1's in [math]\displaystyle{ A }[/math].

Proof. Let [math]\displaystyle{ r }[/math] denote the maximum number of independent 1's in [math]\displaystyle{ A }[/math] and [math]\displaystyle{ s }[/math] be the minimum number of rows and columns to cover all 1's in [math]\displaystyle{ A }[/math]. Clearly, [math]\displaystyle{ r\le s }[/math], since any set of [math]\displaystyle{ r }[/math] independent 1's requires together [math]\displaystyle{ r }[/math] rows and columns to cover. We now prove [math]\displaystyle{ r\ge s }[/math]. Assume that some [math]\displaystyle{ a }[/math] rows and [math]\displaystyle{ b }[/math] columns cover all the 1's in [math]\displaystyle{ A }[/math], and [math]\displaystyle{ s=a+b }[/math], i.e. the covering is minimum. Because permuting the rows or the columns change neither [math]\displaystyle{ r }[/math] nor [math]\displaystyle{ s }[/math] (as reordering the vertices on either side in a bipartite graph changes nothing to the size of matchings and vertex covers), we may assume that the first [math]\displaystyle{ a }[/math] rows and the first [math]\displaystyle{ b }[/math] columns cover the 1's. Write [math]\displaystyle{ A }[/math] in the form [math]\displaystyle{ A=\begin{bmatrix} B_{a\times b} &C_{a\times (n-b)}\\ B_{(m-a)\times b} &E_{(m-a)\times (n-b)} \end{bmatrix} }[/math], where the submatrix [math]\displaystyle{ E }[/math] has only zero entries. We will show that there are [math]\displaystyle{ a }[/math] independent 1's in [math]\displaystyle{ C }[/math] and [math]\displaystyle{ b }[/math] independent 1's in [math]\displaystyle{ D }[/math], thus together [math]\displaystyle{ A }[/math] has [math]\displaystyle{ a+b=s }[/math] independent 1's, which will imply that [math]\displaystyle{ r\ge s }[/math], as desired. Define [math]\displaystyle{ S_i=\{j\mid c_{ij}=1\} }[/math] It is obvious that [math]\displaystyle{ S_i\subseteq\{1,2,\ldots,n-b\} }[/math]. We claim that the sequence [math]\displaystyle{ S_1,S_2,\ldots, S_a }[/math] has a system of distinct representatives, i.e., we can choose a 1 from each row, no two in the same column. Otherwise, Hall's theorem tells us that there exists some [math]\displaystyle{ I\subseteq\{1,2,\ldots,a\} }[/math], such that [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\lt |I| }[/math], that is, the 1's in the rows contained by [math]\displaystyle{ I }[/math] can be covered by less than [math]\displaystyle{ |I| }[/math] columns. Thus, the 1's in [math]\displaystyle{ C }[/math] can be covered by [math]\displaystyle{ a-|I| }[/math] and less than [math]\displaystyle{ |I| }[/math] columns, altogether less than [math]\displaystyle{ a }[/math] rows and columns. Therefore, the 1's in [math]\displaystyle{ A }[/math] can be covered by less than [math]\displaystyle{ a+b }[/math] rows and columns, which contradicts the assumption that the size of the minimum covering of all 1's in [math]\displaystyle{ A }[/math] is [math]\displaystyle{ a+b }[/math]. Therefore, we show that [math]\displaystyle{ C }[/math] has [math]\displaystyle{ a }[/math] independent 1's. By the same argument, we can show that [math]\displaystyle{ D }[/math] has [math]\displaystyle{ b }[/math] independent 1's. Since [math]\displaystyle{ C }[/math] and [math]\displaystyle{ D }[/math] share no rows or columns, the number of independent 1's in [math]\displaystyle{ A }[/math] is [math]\displaystyle{ r\ge a+b=s }[/math]. [math]\displaystyle{ \square }[/math]

Chains and antichains

Dilworth's theorem

Dilworth's Theorem

Suppose that the largest antichain in the poset [math]\displaystyle{ P }[/math] has size [math]\displaystyle{ r }[/math]. Then [math]\displaystyle{ P }[/math] can be partitioned into [math]\displaystyle{ r }[/math] chains.

Proof. Suppose [math]\displaystyle{ P }[/math] has an antichain [math]\displaystyle{ |A| }[/math], and [math]\displaystyle{ P }[/math] can be partitioned into disjoint chains [math]\displaystyle{ C_1,C_2,\ldots,C_s }[/math]. Then [math]\displaystyle{ |A|\le s }[/math], since every chain can pass though an antichain at most once, that is, [math]\displaystyle{ |C_i\cap A|\le 1 }[/math] for all [math]\displaystyle{ i=1,2,\ldots,s }[/math]. Therefore, we only need to prove that there exist an antichain [math]\displaystyle{ A\subseteq P }[/math] of size [math]\displaystyle{ r }[/math], and a partition of [math]\displaystyle{ P }[/math] into at most [math]\displaystyle{ r }[/math] chains. Define a bipartite graph [math]\displaystyle{ G(U,V,E) }[/math] where [math]\displaystyle{ U=V=P }[/math], and for any [math]\displaystyle{ u\in U }[/math] and [math]\displaystyle{ v\in v }[/math], [math]\displaystyle{ uv\in E }[/math] if and only if [math]\displaystyle{ u\lt v }[/math] in the poset [math]\displaystyle{ P }[/math]. By König-Egerváry theorem, there is a matching [math]\displaystyle{ M\subseteq E }[/math] and a vertex set [math]\displaystyle{ C }[/math] such that every edge in [math]\displaystyle{ E }[/math] is adjacent to at least a vertex in [math]\displaystyle{ C }[/math], and [math]\displaystyle{ |M|=|C| }[/math]. Denote [math]\displaystyle{ |M|=|C|=m }[/math]. Let [math]\displaystyle{ A }[/math] be the set of uncovered elements in poset [math]\displaystyle{ P }[/math], i.e., the elements of [math]\displaystyle{ P }[/math] that do not correspond to any vertex in [math]\displaystyle{ C }[/math]. Clearly, [math]\displaystyle{ |A|\ge n-m }[/math]. We claim that [math]\displaystyle{ A\subseteq P }[/math] is an antichain. By contradiction, assume there exists [math]\displaystyle{ x,y\in A }[/math] such that [math]\displaystyle{ x\lt y }[/math]. Then, by the definition of [math]\displaystyle{ G }[/math], there exist [math]\displaystyle{ u_x\in U,v_x\in V }[/math] which corresponds to [math]\displaystyle{ x }[/math], and [math]\displaystyle{ u_y\in U,v_y\in V }[/math] which corresponds to [math]\displaystyle{ y }[/math], such that [math]\displaystyle{ u_xv_y\in E }[/math]. But since [math]\displaystyle{ A }[/math] includes only those elements whose corresponding vertices are not in [math]\displaystyle{ C }[/math], none of [math]\displaystyle{ u_x,v_x,u_y,v_y }[/math] can be in [math]\displaystyle{ C }[/math], which contradicts the fact that [math]\displaystyle{ C }[/math] is a vertex cover of [math]\displaystyle{ G }[/math] that every edges in [math]\displaystyle{ G }[/math] are adjacent to at least a vertex in [math]\displaystyle{ C }[/math]. Let [math]\displaystyle{ B }[/math] be a family of chains formed by including [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] in the same chain whenever [math]\displaystyle{ uv\in M }[/math]. A moment thought would tell us that the number of chains in [math]\displaystyle{ B }[/math] is equal to the unmatched vertices in [math]\displaystyle{ U }[/math] (or [math]\displaystyle{ V }[/math]). Thus, [math]\displaystyle{ |B|=n-m }[/math]. Altogether, we construct an antichain of size [math]\displaystyle{ |A|\ge n-m }[/math] and partition the poset [math]\displaystyle{ P }[/math] into [math]\displaystyle{ |B|=n-m }[/math] disjoint chains. The theorem is proved. [math]\displaystyle{ \square }[/math]

Application: Erdős-Szekeres Theorem

Erdős-Szekeres Theorem

A sequence of more than [math]\displaystyle{ mn }[/math] different real numbers must contain either an increasing subsequence of length [math]\displaystyle{ m+1 }[/math], or a decreasing subsequence of length [math]\displaystyle{ n+1 }[/math].

Proof by Dilworth's theorem (Original proof of Erdős-Szekeres) Let [math]\displaystyle{ (a_1,a_2,\ldots,a_N) }[/math] be the sequence of [math]\displaystyle{ N\gt mn }[/math] distinct real numbers. Define the poset as [math]\displaystyle{ P=\{(i,a_i)\mid i=1,2,\ldots,N\} }[/math] and [math]\displaystyle{ (i,a_i)\le (j,a_j) }[/math] if and only if [math]\displaystyle{ i\le j }[/math] and [math]\displaystyle{ a_i\le a_j }[/math]. A chain [math]\displaystyle{ (i_1,a_{i_1})\lt (i_2,a_{i_2})\lt \cdots\lt (i_k,a_{i_k}) }[/math] must have [math]\displaystyle{ i_1\lt i_2\lt \cdots\lt i_k }[/math] and [math]\displaystyle{ a_{i_1}\lt a_{i_2}\lt \cdots\lt a_{i_k} }[/math]. Thus, each chain correspond to an increasing subsequence. Let [math]\displaystyle{ (j_1,a_{j_1}),(j_2,a_{j_2}),\cdots,(j_k,a_{j_k}) }[/math] be an antichain. Without loss of generality, we can assume that [math]\displaystyle{ j_1\lt j_2\lt \cdots\lt j_k }[/math]. The only case that these elements are non-comparable is that [math]\displaystyle{ a_{j_1}\gt a_{j_2}\gt \cdots\gt a_{j_k} }[/math], otherwise if [math]\displaystyle{ a_{j_s}\lt a_{j_t} }[/math] for some [math]\displaystyle{ s\lt t }[/math], then [math]\displaystyle{ (j_s,a_{j_s})\lt (j_t,a_{j_t}) }[/math], which contradicts the fact that it is an antichain. Thus, each antichain corresponds to a decreasing subsequence. If [math]\displaystyle{ P }[/math] has an antichain of size [math]\displaystyle{ n+1 }[/math], then [math]\displaystyle{ (a_1,a_2,\ldots,a_N) }[/math] has a decreasing subsequence of size [math]\displaystyle{ n+1 }[/math], and we are done. Alternatively, if the largest antichain in [math]\displaystyle{ P }[/math] is of size at most [math]\displaystyle{ n }[/math], then by Dilworth's theorem, [math]\displaystyle{ P }[/math] can be partitioned into at most [math]\displaystyle{ n }[/math] disjoint chains, due to pigeonhole principle, one of which must be of length [math]\displaystyle{ m+1 }[/math], which means [math]\displaystyle{ (a_1,a_2,\ldots,a_N) }[/math] has an increasing subsequence of size [math]\displaystyle{ m+1 }[/math]. [math]\displaystyle{ \square }[/math]

Application: Hall's Theorem

Hall's Theorem

The sets [math]\displaystyle{ S_1,S_2,\ldots,S_m }[/math] have a system of distinct representatives (SDR) if and only if [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\ge |I| }[/math] for all [math]\displaystyle{ I\subseteq\{1,2,\ldots,m\} }[/math].

Proof by Dilworth's theorem As we discussed before, the necessity of Hall's condition for the existence of SDR is easy. We prove its sufficiency by Dilworth's theorem. Denote [math]\displaystyle{ X=\bigcup_{i=1}^mS_i }[/math]. Construct a poset [math]\displaystyle{ P }[/math] by letting [math]\displaystyle{ P=X\cup\{S_1,S_2,\ldots,S_m\} }[/math] and [math]\displaystyle{ x\lt S_{i} }[/math] for all [math]\displaystyle{ x\in S_{i} }[/math]. There are no other comparabilities. It is obvious that [math]\displaystyle{ X }[/math] is an antichain. We claim it is also the largest one. To prove this, let [math]\displaystyle{ A }[/math] be an arbitrary antichain, and let [math]\displaystyle{ I=\{i\mid S_i\in A\} }[/math]. Then [math]\displaystyle{ A }[/math] contains no elements of [math]\displaystyle{ \bigcup_{i\in I}S_i }[/math], since if [math]\displaystyle{ x\in A }[/math] and [math]\displaystyle{ x\in S_i\in A }[/math], then [math]\displaystyle{ x\lt S_i }[/math] and [math]\displaystyle{ A }[/math] cannot be an antichain. Thus, [math]\displaystyle{ |A|\le |I|+|X|-\left|\bigcup_{i\in I}S_i\right| }[/math] and by Hall's condition [math]\displaystyle{ \left|\bigcup_{i\in I}S_i\right|\ge|I| }[/math], thus [math]\displaystyle{ |A|\le |X| }[/math], as claimed. Now, Dilworth's theorem implies that [math]\displaystyle{ P }[/math] can be partitioned into [math]\displaystyle{ |X| }[/math] chains. Since [math]\displaystyle{ X }[/math] is an antichain and each chain can pass though an antichain on at most one element, each of the [math]\displaystyle{ |X| }[/math] chains contain precisely one [math]\displaystyle{ x\in X }[/math]. And since [math]\displaystyle{ \{S_1,\ldots,S_m\} }[/math] is also an antichain, each of these [math]\displaystyle{ |X| }[/math] chains contain at most one [math]\displaystyle{ S_i }[/math]. Altogether, the [math]\displaystyle{ |X| }[/math] chains are in the form: [math]\displaystyle{ \{x_1,S_1\},\{x_2,S_2\},\ldots,\{x_m,S_m\},\{x_{m+1}\}\ldots,\{x_{|X|}\} }[/math]. Since the only comparabilities in our posets are [math]\displaystyle{ x\in S_i }[/math] and the above chains are disjoint, we have [math]\displaystyle{ x_1\in S_1, x_2\in S_2,\ldots,x_m\in S_m }[/math] as an SDR. [math]\displaystyle{ \square }[/math]