Clarification regarding Dilworth Theorem Proof

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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This is the proof I am talking about.



http://www.math.cmu.edu/~af1p/Teaching/Combinatorics/F03/Class14.pdf



It is given that :
P⁻∩ P⁺=A
Otherwise there exists x,i, j such that ai < x < aj and so A is not an
anti-chain.



Please could anybody explain this to me ? Why should the intersection of P⁻ and P⁺ be equal to A ?




combinatorics extremal-combinatorics




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  • Can you explain what $P^pm$, $A$ etc. are without having to refer us to this pdf?
    – Lord Shark the Unknown
    Aug 27 at 15:37










  • A is an antichain A = a1, a2, . . . , am in P C. C is a maximal chain in the poset P . P⁺ = x ∈ P : x ≤ ai for some i. P⁻ = x ∈ P : x ≥ ai for some i.
    – Arka Prava Paul
    Aug 27 at 15:48














up vote
0
down vote

favorite
1












This is the proof I am talking about.



http://www.math.cmu.edu/~af1p/Teaching/Combinatorics/F03/Class14.pdf



It is given that :
P⁻∩ P⁺=A
Otherwise there exists x,i, j such that ai < x < aj and so A is not an
anti-chain.



Please could anybody explain this to me ? Why should the intersection of P⁻ and P⁺ be equal to A ?




combinatorics extremal-combinatorics




share|cite|improve this question




















  • Can you explain what $P^pm$, $A$ etc. are without having to refer us to this pdf?
    – Lord Shark the Unknown
    Aug 27 at 15:37










  • A is an antichain A = a1, a2, . . . , am in P C. C is a maximal chain in the poset P . P⁺ = x ∈ P : x ≤ ai for some i. P⁻ = x ∈ P : x ≥ ai for some i.
    – Arka Prava Paul
    Aug 27 at 15:48












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





This is the proof I am talking about.



http://www.math.cmu.edu/~af1p/Teaching/Combinatorics/F03/Class14.pdf



It is given that :
P⁻∩ P⁺=A
Otherwise there exists x,i, j such that ai < x < aj and so A is not an
anti-chain.



Please could anybody explain this to me ? Why should the intersection of P⁻ and P⁺ be equal to A ?




combinatorics extremal-combinatorics




share|cite|improve this question












This is the proof I am talking about.



http://www.math.cmu.edu/~af1p/Teaching/Combinatorics/F03/Class14.pdf



It is given that :
P⁻∩ P⁺=A
Otherwise there exists x,i, j such that ai < x < aj and so A is not an
anti-chain.



Please could anybody explain this to me ? Why should the intersection of P⁻ and P⁺ be equal to A ?





combinatorics extremal-combinatorics





share|cite|improve this question











share|cite|improve this question




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asked Aug 27 at 15:35









Arka Prava Paul

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  • Can you explain what $P^pm$, $A$ etc. are without having to refer us to this pdf?
    – Lord Shark the Unknown
    Aug 27 at 15:37










  • A is an antichain A = a1, a2, . . . , am in P C. C is a maximal chain in the poset P . P⁺ = x ∈ P : x ≤ ai for some i. P⁻ = x ∈ P : x ≥ ai for some i.
    – Arka Prava Paul
    Aug 27 at 15:48
















  • Can you explain what $P^pm$, $A$ etc. are without having to refer us to this pdf?
    – Lord Shark the Unknown
    Aug 27 at 15:37










  • A is an antichain A = a1, a2, . . . , am in P C. C is a maximal chain in the poset P . P⁺ = x ∈ P : x ≤ ai for some i. P⁻ = x ∈ P : x ≥ ai for some i.
    – Arka Prava Paul
    Aug 27 at 15:48















Can you explain what $P^pm$, $A$ etc. are without having to refer us to this pdf?
– Lord Shark the Unknown
Aug 27 at 15:37




Can you explain what $P^pm$, $A$ etc. are without having to refer us to this pdf?
– Lord Shark the Unknown
Aug 27 at 15:37












A is an antichain A = a1, a2, . . . , am in P C. C is a maximal chain in the poset P . P⁺ = x ∈ P : x ≤ ai for some i. P⁻ = x ∈ P : x ≥ ai for some i.
– Arka Prava Paul
Aug 27 at 15:48




A is an antichain A = a1, a2, . . . , am in P C. C is a maximal chain in the poset P . P⁺ = x ∈ P : x ≤ ai for some i. P⁻ = x ∈ P : x ≥ ai for some i.
– Arka Prava Paul
Aug 27 at 15:48










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Since $A$ is obviously included in $P^-$ and in $P^+$, we have $Asubseteq P^-cap P^+$. The problem is to prove the reverse inclusion. So consider any element $xin P^-cap P^+$; I need to show that $xin A$. By definition of $P^-$ there is an $i$ such that $xleq a_i$. If $x=a_i$, that means $xin A$, so I'm done. It remains to consider the situation when $x<a_i$. Similarly, by definition of $P^+$, $xgeq a_j$ for some $j$, and I'm done if equality holds, so I need only consider the situation when $x>a_j$. But then $a_j<x<a_i$, which means that two different elements of $A$ are comparable: $a_j<a_i$. That contradicts the fact that $A$ is an antichain.






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    Since $A$ is obviously included in $P^-$ and in $P^+$, we have $Asubseteq P^-cap P^+$. The problem is to prove the reverse inclusion. So consider any element $xin P^-cap P^+$; I need to show that $xin A$. By definition of $P^-$ there is an $i$ such that $xleq a_i$. If $x=a_i$, that means $xin A$, so I'm done. It remains to consider the situation when $x<a_i$. Similarly, by definition of $P^+$, $xgeq a_j$ for some $j$, and I'm done if equality holds, so I need only consider the situation when $x>a_j$. But then $a_j<x<a_i$, which means that two different elements of $A$ are comparable: $a_j<a_i$. That contradicts the fact that $A$ is an antichain.






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      up vote
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      down vote













      Since $A$ is obviously included in $P^-$ and in $P^+$, we have $Asubseteq P^-cap P^+$. The problem is to prove the reverse inclusion. So consider any element $xin P^-cap P^+$; I need to show that $xin A$. By definition of $P^-$ there is an $i$ such that $xleq a_i$. If $x=a_i$, that means $xin A$, so I'm done. It remains to consider the situation when $x<a_i$. Similarly, by definition of $P^+$, $xgeq a_j$ for some $j$, and I'm done if equality holds, so I need only consider the situation when $x>a_j$. But then $a_j<x<a_i$, which means that two different elements of $A$ are comparable: $a_j<a_i$. That contradicts the fact that $A$ is an antichain.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Since $A$ is obviously included in $P^-$ and in $P^+$, we have $Asubseteq P^-cap P^+$. The problem is to prove the reverse inclusion. So consider any element $xin P^-cap P^+$; I need to show that $xin A$. By definition of $P^-$ there is an $i$ such that $xleq a_i$. If $x=a_i$, that means $xin A$, so I'm done. It remains to consider the situation when $x<a_i$. Similarly, by definition of $P^+$, $xgeq a_j$ for some $j$, and I'm done if equality holds, so I need only consider the situation when $x>a_j$. But then $a_j<x<a_i$, which means that two different elements of $A$ are comparable: $a_j<a_i$. That contradicts the fact that $A$ is an antichain.






        share|cite|improve this answer












        Since $A$ is obviously included in $P^-$ and in $P^+$, we have $Asubseteq P^-cap P^+$. The problem is to prove the reverse inclusion. So consider any element $xin P^-cap P^+$; I need to show that $xin A$. By definition of $P^-$ there is an $i$ such that $xleq a_i$. If $x=a_i$, that means $xin A$, so I'm done. It remains to consider the situation when $x<a_i$. Similarly, by definition of $P^+$, $xgeq a_j$ for some $j$, and I'm done if equality holds, so I need only consider the situation when $x>a_j$. But then $a_j<x<a_i$, which means that two different elements of $A$ are comparable: $a_j<a_i$. That contradicts the fact that $A$ is an antichain.







        share|cite|improve this answer












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        answered Aug 27 at 18:49









        Andreas Blass

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