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CS 207 Discrete Mathematics – 2012-2013
Nutan Limaye
Indian Institute of Technology, [email protected]
Mathematical Reasoning and Mathematical ObjectsLecture 7: Properties of equivalence relations and partial orders
August 13, 2012
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Last time
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Recap
What are relaions?
What are different types of functions?reflexive, transitive, symmetric, anti-symmetric
Equivalence relations and partial orders.reflexive transitive symmetric anti-symmetric
Representation of partial orders by graphs
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Recap
What are relaions?
What are different types of functions?
reflexive, transitive, symmetric, anti-symmetric
Equivalence relations and partial orders.reflexive transitive symmetric anti-symmetric
equivalence X X Xrelation
Representation of partial orders by graphs
Nutan (IITB) CS 207 Discrete Mathematics – 2012-2013 May 2011 3 / 14
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Recap
What are relaions?
What are different types of functions?reflexive, transitive, symmetric, anti-symmetric
Equivalence relations and partial orders.reflexive transitive symmetric anti-symmetric
equivalence X X Xrelationpartial order X X X
Representation of partial orders by graphs
Nutan (IITB) CS 207 Discrete Mathematics – 2012-2013 May 2011 3 / 14
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Recap
What are relaions?
What are different types of functions?reflexive, transitive, symmetric, anti-symmetric
Equivalence relations and partial orders.
reflexive transitive symmetric anti-symmetric
equivalence X X Xrelationpartial order X X X
Representation of partial orders by graphs
Nutan (IITB) CS 207 Discrete Mathematics – 2012-2013 May 2011 3 / 14
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Recap
What are relaions?
What are different types of functions?reflexive, transitive, symmetric, anti-symmetric
Equivalence relations and partial orders.reflexive transitive symmetric anti-symmetric
equivalence X X Xrelationpartial order X X X
Representation of partial orders by graphs
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Today
What are equivalence classes and properties of equivalence classes.
Recall chains and anti-chains and study properties of partial orders.
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Today
What are equivalence classes and properties of equivalence classes.
Recall chains and anti-chains and study properties of partial orders.
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Equivalence relations and equivalence classes
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called anequivalence relation if it is reflexive, transitive and symmetric.
Definition
Let [x ] := {y | x , y ∈ A, and (x , y) ∈ R}.[x ] is called the equivalence class of x .
Example: Consider (N,≡ (mod4)).
[0] = {0, 4, 8, 12, 16, . . .}[1] = {1, 5, 9, 13, 17, . . .}
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Equivalence relations and equivalence classes
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called anequivalence relation if it is reflexive, transitive and symmetric.
Definition
Let [x ] := {y | x , y ∈ A, and (x , y) ∈ R}.[x ] is called the equivalence class of x .
Example: Consider (N,≡ (mod4)).
[0]
= {0, 4, 8, 12, 16, . . .}[1] = {1, 5, 9, 13, 17, . . .}
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Equivalence relations and equivalence classes
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called anequivalence relation if it is reflexive, transitive and symmetric.
Definition
Let [x ] := {y | x , y ∈ A, and (x , y) ∈ R}.[x ] is called the equivalence class of x .
Example: Consider (N,≡ (mod4)).
[0] = {0, 4, 8, 12, 16, . . .}
[1] = {1, 5, 9, 13, 17, . . .}
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Equivalence relations and equivalence classes
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called anequivalence relation if it is reflexive, transitive and symmetric.
Definition
Let [x ] := {y | x , y ∈ A, and (x , y) ∈ R}.[x ] is called the equivalence class of x .
Example: Consider (N,≡ (mod4)).
[0] = {0, 4, 8, 12, 16, . . .}[1]
= {1, 5, 9, 13, 17, . . .}
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Equivalence relations and equivalence classes
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called anequivalence relation if it is reflexive, transitive and symmetric.
Definition
Let [x ] := {y | x , y ∈ A, and (x , y) ∈ R}.[x ] is called the equivalence class of x .
Example: Consider (N,≡ (mod4)).
[0] = {0, 4, 8, 12, 16, . . .}[1] = {1, 5, 9, 13, 17, . . .}
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Properties of equivalence relations
Let R be an equivalence relation of A. Let elements of A be x , y , z etc.
Lemma
The following three are equivalent: (a) xRy, (b) [x ] = [y ], (c) [x ] ∩ [y ] 6= ∅.
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Properties of equivalence relations
Let R be an equivalence relation of A. Let elements of A be x , y , z etc.
Lemma
The following three are equivalent: (a) xRy, (b) [x ] = [y ], (c) [x ] ∩ [y ] 6= ∅.
Proof.
(a) ⇒ (b): Say z ∈ [x ]. But xRy . As xRy and R is symmetric, yRx .Therefore, yRx , xRz . R is transitive. Therefore, yRz , i.e. z ∈ [y ]. Thisproves that [x ] ⊆ [y ]. The proof of [y ] ⊆ [x ] is similar.
(b) ⇒ (c): Say [x ] = [y ]. The only way [x ] ∩ [y ] = ∅ is if [x ] = ∅.However, as R is reflexive, x ∈ [x ] 6= ∅.(c) ⇒ (a): Let z ∈ [x ] ∩ [y ]. Therefore, xRz and yRz . But as R issymmetric, zRy . But R is also transitive. Therefore xRz and zRy implyxRy .
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Properties of equivalence relations
Let R be an equivalence relation of A. Let elements of A be x , y , z etc.
Lemma
The following three are equivalent: (a) xRy, (b) [x ] = [y ], (c) [x ] ∩ [y ] 6= ∅.
Proof.
(a) ⇒ (b): Say z ∈ [x ]. But xRy . As xRy and R is symmetric, yRx .Therefore, yRx , xRz . R is transitive. Therefore, yRz , i.e. z ∈ [y ]. Thisproves that [x ] ⊆ [y ]. The proof of [y ] ⊆ [x ] is similar.(b) ⇒ (c): Say [x ] = [y ]. The only way [x ] ∩ [y ] = ∅ is if [x ] = ∅.However, as R is reflexive, x ∈ [x ] 6= ∅.
(c) ⇒ (a): Let z ∈ [x ] ∩ [y ]. Therefore, xRz and yRz . But as R issymmetric, zRy . But R is also transitive. Therefore xRz and zRy implyxRy .
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Properties of equivalence relations
Let R be an equivalence relation of A. Let elements of A be x , y , z etc.
Lemma
The following three are equivalent: (a) xRy, (b) [x ] = [y ], (c) [x ] ∩ [y ] 6= ∅.
Proof.
(a) ⇒ (b): Say z ∈ [x ]. But xRy . As xRy and R is symmetric, yRx .Therefore, yRx , xRz . R is transitive. Therefore, yRz , i.e. z ∈ [y ]. Thisproves that [x ] ⊆ [y ]. The proof of [y ] ⊆ [x ] is similar.(b) ⇒ (c): Say [x ] = [y ]. The only way [x ] ∩ [y ] = ∅ is if [x ] = ∅.However, as R is reflexive, x ∈ [x ] 6= ∅.(c) ⇒ (a): Let z ∈ [x ] ∩ [y ]. Therefore, xRz and yRz . But as R issymmetric, zRy . But R is also transitive. Therefore xRz and zRy implyxRy .
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Sets X1,X2, . . . ,Xm are said to partition a set X if
∀i , j ∈ {1, 2, . . . ,m}, i 6= j : Xi ∩ Xj = ∅∀x ∈ X , ∃i ∈ {1, 2, . . . ,m} : x ∈ Xi
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅.
Also, for each x ∈ A, x ∈ [x ].
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Conversely, given a partition {Ai | i ∈ {1, 2, . . . , n}} of A, there is anequivalence relation RA with equivalence classes A1,A2, . . . ,An.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Conversely, given a partition {Ai | i ∈ {1, 2, . . . , n}} of A, there is anequivalence relation RA with equivalence classes A1,A2, . . . ,An.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Conversely, given a partition {Ai | i ∈ {1, 2, . . . , n}} of A, there is anequivalence relation RA with equivalence classes A1,A2, . . . ,An.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
Let RA = {(x , y) | ∃i : x , y ∈ Ai}.
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Conversely, given a partition {Ai | i ∈ {1, 2, . . . , n}} of A, there is anequivalence relation RA with equivalence classes A1,A2, . . . ,An.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
Let RA = {(x , y) | ∃i : x , y ∈ Ai}.RA relates (x , y) if they belong to the same part in the partition of A.
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Equivalence classes and partitions
Theorem
Let R be an equivalence relation defined on a set A.
The equivalence classes of R, partition the set A.
Conversely, given a partition {Ai | i ∈ {1, 2, . . . , n}} of A, there is anequivalence relation RA with equivalence classes A1,A2, . . . ,An.
Proof.
Let [x ] 6= [y ] be two distinct equivalence classes of R. From the previouslemma [x ] ∩ [y ] = ∅. Also, for each x ∈ A, x ∈ [x ].
Let RA = {(x , y) | ∃i : x , y ∈ Ai}.RA is reflexive. If (x , y) ∈ RA then even (y , x) ∈ RA. Finally, if (x , y) ∈ RA
then ∃i : x , y ∈ Ai . Let that index be called i0. Now if (y , z) ∈ RA thenboth y , z must be in the same part of the partition. But we know thaty ∈ Ai0 . Therefore, z ∈ Ai0 . Hence, x , z ∈ Ai0 and hence (x , z) ∈ RA. Thisproves that RA is also transitive.
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Partial orders, chains, anti-chains
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called apartially ordered set or a poset if it is reflexive, transitive andanti-symmetric.
Definition
If (S ,�) is a poset and A ⊆ S such that every pair of elements in A iscomparable as per �, then A is called a chain.
Definition
Let (S ,�) be a poset. A subset A ⊆ S is called an anti-chain if no twoelements of A are related to reach other under �.
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Partial orders, chains, anti-chains
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called apartially ordered set or a poset if it is reflexive, transitive andanti-symmetric.
Definition
If (S ,�) is a poset and A ⊆ S such that every pair of elements in A iscomparable as per �, then A is called a chain.
Definition
Let (S ,�) be a poset. A subset A ⊆ S is called an anti-chain if no twoelements of A are related to reach other under �.
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Partial orders, chains, anti-chains
Definition
A relation R defined over a set A, denoted as R(A) or (A,R), is called apartially ordered set or a poset if it is reflexive, transitive andanti-symmetric.
Definition
If (S ,�) is a poset and A ⊆ S such that every pair of elements in A iscomparable as per �, then A is called a chain.
Definition
Let (S ,�) be a poset. A subset A ⊆ S is called an anti-chain if no twoelements of A are related to reach other under �.
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Example
Let S = {1, 2, 3}. Recall the poset (S ,⊆).
∅
{1} {2} {3}
{1, 2} {1, 3} {2, 3}
{1, 2, 3}
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Example
Let S = {1, 2, 3}. Recall the poset (S ,⊆).
∅
{1} {2} {3}
{1, 2} {1, 3} {2, 3}
{1, 2, 3}
Nutan (IITB) CS 207 Discrete Mathematics – 2012-2013 May 2011 9 / 14
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Chains and anti-chains
Theorem
If the largest chain in a poset (S ,�) is of size m then S has at least manti-chains.
Proof.
Let the chain be denoted as a1 � a2 � . . . � am. Now observe that everyelement of this chain, must go to different anti-chains. Therefore, thereare at least m anti-chains in (S ,�).
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Chains and anti-chains
Theorem
If the largest chain in a poset (S ,�) is of size m then S has at least manti-chains. The size of an anti-chain is the number of elements in thechain.
Proof.
Let the chain be denoted as a1 � a2 � . . . � am. Now observe that everyelement of this chain, must go to different anti-chains. Therefore, thereare at least m anti-chains in (S ,�).
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Chains and anti-chains
Theorem
If the largest chain in a poset (S ,�) is of size m then S has at least manti-chains.
Proof.
Let the chain be denoted as a1 � a2 � . . . � am. Now observe that everyelement of this chain, must go to different anti-chains.
Therefore, thereare at least m anti-chains in (S ,�).
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Chains and anti-chains
Theorem
If the largest chain in a poset (S ,�) is of size m then S has at least manti-chains.
Proof.
Let the chain be denoted as a1 � a2 � . . . � am. Now observe that everyelement of this chain, must go to different anti-chains. Therefore, thereare at least m anti-chains in (S ,�).
Nutan (IITB) CS 207 Discrete Mathematics – 2012-2013 May 2011 10 / 14
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.
Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.[CW] For any s ∈ S , how large can label(s) be?
Let us now define setsA1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.
It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .
Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m].
∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. [m] = {1, 2, . . . ,m}
∴ label(x) = label(y) = i . Suppose x � ythen label(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction!
Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction.
Hence, every Ai is an anti-chain.
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Chains and anti-chains
Theorem (Mirsky’s theorem, 1971)
If the largest chain in a poset (S ,�) is of size m then S can be partitionedinto m anti-chains.
Proof.
For each element s ∈ S , let Cs be the set of all chains that have s as themaximum element. And define label(s) := maxc∈Cs {size(c)}.Let us now define sets A1,A2, . . . ,Am such that Ai = {x | label(x) = i}.It is easy to see that if i 6= j then Ai ∩ Aj = ∅. Also, it is easy to observethat ∪mi=1Ai = S .Now we prove that each Ai is an anti-chain. For x , y ∈ Ai for somei ∈ [m]. ∴ label(x) = label(y) = i . Suppose x � y thenlabel(x) < label(y). Contradiction! Similarly, if x � y then we get acontradiction. Hence, every Ai is an anti-chain.
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Subset Take-away problem
Removing supersets ≡ getting rid of some chains.
Do partial orders help?
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Subset Take-away problem
Removing supersets ≡ getting rid of some chains.Do partial orders help?
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