Maximum cardinality of a set of subsets
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Let $N$ be a system of subsets of the set $X = {1,2,3,cdots ,n }$ such that there are no three elements $A,B,C in N$ such that $A subset B subset C$. Prove that $$|N| leq 2 cdot {{n}choose{ lfloor n/2 rfloor}}. $$
I have thought about the fact that a semi-independent system of subsets can have at most two elements in common with any chain from $(mathcal{P}(X), subseteq),$ but I do not know how to continue using Sperner's theorem.
combinatorics discrete-mathematics order-theory extremal-combinatorics
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Let $N$ be a system of subsets of the set $X = {1,2,3,cdots ,n }$ such that there are no three elements $A,B,C in N$ such that $A subset B subset C$. Prove that $$|N| leq 2 cdot {{n}choose{ lfloor n/2 rfloor}}. $$
I have thought about the fact that a semi-independent system of subsets can have at most two elements in common with any chain from $(mathcal{P}(X), subseteq),$ but I do not know how to continue using Sperner's theorem.
combinatorics discrete-mathematics order-theory extremal-combinatorics
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add a comment |
$begingroup$
Let $N$ be a system of subsets of the set $X = {1,2,3,cdots ,n }$ such that there are no three elements $A,B,C in N$ such that $A subset B subset C$. Prove that $$|N| leq 2 cdot {{n}choose{ lfloor n/2 rfloor}}. $$
I have thought about the fact that a semi-independent system of subsets can have at most two elements in common with any chain from $(mathcal{P}(X), subseteq),$ but I do not know how to continue using Sperner's theorem.
combinatorics discrete-mathematics order-theory extremal-combinatorics
$endgroup$
Let $N$ be a system of subsets of the set $X = {1,2,3,cdots ,n }$ such that there are no three elements $A,B,C in N$ such that $A subset B subset C$. Prove that $$|N| leq 2 cdot {{n}choose{ lfloor n/2 rfloor}}. $$
I have thought about the fact that a semi-independent system of subsets can have at most two elements in common with any chain from $(mathcal{P}(X), subseteq),$ but I do not know how to continue using Sperner's theorem.
combinatorics discrete-mathematics order-theory extremal-combinatorics
combinatorics discrete-mathematics order-theory extremal-combinatorics
edited Jan 12 at 22:33
asked Jan 12 at 17:49
user606835
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2 Answers
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Let $mathcal{M}$ be the family of all maximal chains. (We call a chain ${varnothingsubsetneq A_1subsetneq A_2subsetneqcdots subsetneq A_n=X}subset 2^X$ of length $n+1$ as a maximal chain.) Define
$$
Q=sum_{Min mathcal{M}}sum_{Ain N} 1_{Ain M}
$$ where $1_{Ain M}=1$ if $Ain M$ and is $0$ otherwise. Since for every maximal chain $Minmathcal{M}$, there are at most $2$ distinct $Ain N$ such that $Ain M$, we have
$$
Qlesum_{Min mathcal{M}}2=2|mathcal{M}|=2cdot n!.
$$On the other hand, we have
$$
Q=sum_{Ain N}sum_{Min mathcal{M}}1_{Ain M}=sum_{Ain N}|A|!(n-|A|)!
$$ since the number of maximal chains containing $A$ is $|A|!(n-|A|)!$. Note that $j!(n-j)!$ is minimized when $j=lfloor frac{n}{2}rfloor$. Thus, we have
$$
Q= sum_{Ain N}|A|!(n-|A|)!ge lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|
$$ Gathering them together, we get
$$
lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|le 2cdot n!
$$or
$$
|N|le 2cdot frac{n!}{lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!}=2cdot binom{n}{lfloor frac{n}{2}rfloor},
$$ as desired.
answered Jan 12 at 18:09
SongSong
18.6k21651
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If $X$ does not contain a $3$-elament chain $Asubset Bsubset C$, then $X$ is the union of two antichains. Namely, the set of all minimal elements of $X$ is an antichain, and the set of all non-minimal elements of $X$ is another antichain. By Sperner's theorem, an antichain of subsets of ${1,2,3,dots,n}$ has at most $binom n{lfloor n/2rfloor}$ elements, so your set $X$ has at most twice that many elements.
answered Jan 13 at 3:16
bofbof
52.6k559121
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2 Answers
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2 Answers
2
active
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active
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active
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$begingroup$
Let $mathcal{M}$ be the family of all maximal chains. (We call a chain ${varnothingsubsetneq A_1subsetneq A_2subsetneqcdots subsetneq A_n=X}subset 2^X$ of length $n+1$ as a maximal chain.) Define
$$
Q=sum_{Min mathcal{M}}sum_{Ain N} 1_{Ain M}
$$ where $1_{Ain M}=1$ if $Ain M$ and is $0$ otherwise. Since for every maximal chain $Minmathcal{M}$, there are at most $2$ distinct $Ain N$ such that $Ain M$, we have
$$
Qlesum_{Min mathcal{M}}2=2|mathcal{M}|=2cdot n!.
$$On the other hand, we have
$$
Q=sum_{Ain N}sum_{Min mathcal{M}}1_{Ain M}=sum_{Ain N}|A|!(n-|A|)!
$$ since the number of maximal chains containing $A$ is $|A|!(n-|A|)!$. Note that $j!(n-j)!$ is minimized when $j=lfloor frac{n}{2}rfloor$. Thus, we have
$$
Q= sum_{Ain N}|A|!(n-|A|)!ge lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|
$$ Gathering them together, we get
$$
lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|le 2cdot n!
$$or
$$
|N|le 2cdot frac{n!}{lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!}=2cdot binom{n}{lfloor frac{n}{2}rfloor},
$$ as desired.
answered Jan 12 at 18:09
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Let $mathcal{M}$ be the family of all maximal chains. (We call a chain ${varnothingsubsetneq A_1subsetneq A_2subsetneqcdots subsetneq A_n=X}subset 2^X$ of length $n+1$ as a maximal chain.) Define
$$
Q=sum_{Min mathcal{M}}sum_{Ain N} 1_{Ain M}
$$ where $1_{Ain M}=1$ if $Ain M$ and is $0$ otherwise. Since for every maximal chain $Minmathcal{M}$, there are at most $2$ distinct $Ain N$ such that $Ain M$, we have
$$
Qlesum_{Min mathcal{M}}2=2|mathcal{M}|=2cdot n!.
$$On the other hand, we have
$$
Q=sum_{Ain N}sum_{Min mathcal{M}}1_{Ain M}=sum_{Ain N}|A|!(n-|A|)!
$$ since the number of maximal chains containing $A$ is $|A|!(n-|A|)!$. Note that $j!(n-j)!$ is minimized when $j=lfloor frac{n}{2}rfloor$. Thus, we have
$$
Q= sum_{Ain N}|A|!(n-|A|)!ge lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|
$$ Gathering them together, we get
$$
lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|le 2cdot n!
$$or
$$
|N|le 2cdot frac{n!}{lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!}=2cdot binom{n}{lfloor frac{n}{2}rfloor},
$$ as desired.
answered Jan 12 at 18:09
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Let $mathcal{M}$ be the family of all maximal chains. (We call a chain ${varnothingsubsetneq A_1subsetneq A_2subsetneqcdots subsetneq A_n=X}subset 2^X$ of length $n+1$ as a maximal chain.) Define
$$
Q=sum_{Min mathcal{M}}sum_{Ain N} 1_{Ain M}
$$ where $1_{Ain M}=1$ if $Ain M$ and is $0$ otherwise. Since for every maximal chain $Minmathcal{M}$, there are at most $2$ distinct $Ain N$ such that $Ain M$, we have
$$
Qlesum_{Min mathcal{M}}2=2|mathcal{M}|=2cdot n!.
$$On the other hand, we have
$$
Q=sum_{Ain N}sum_{Min mathcal{M}}1_{Ain M}=sum_{Ain N}|A|!(n-|A|)!
$$ since the number of maximal chains containing $A$ is $|A|!(n-|A|)!$. Note that $j!(n-j)!$ is minimized when $j=lfloor frac{n}{2}rfloor$. Thus, we have
$$
Q= sum_{Ain N}|A|!(n-|A|)!ge lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|
$$ Gathering them together, we get
$$
lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|le 2cdot n!
$$or
$$
|N|le 2cdot frac{n!}{lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!}=2cdot binom{n}{lfloor frac{n}{2}rfloor},
$$ as desired.
answered Jan 12 at 18:09
SongSong
18.6k21651
$endgroup$
Let $mathcal{M}$ be the family of all maximal chains. (We call a chain ${varnothingsubsetneq A_1subsetneq A_2subsetneqcdots subsetneq A_n=X}subset 2^X$ of length $n+1$ as a maximal chain.) Define
$$
Q=sum_{Min mathcal{M}}sum_{Ain N} 1_{Ain M}
$$ where $1_{Ain M}=1$ if $Ain M$ and is $0$ otherwise. Since for every maximal chain $Minmathcal{M}$, there are at most $2$ distinct $Ain N$ such that $Ain M$, we have
$$
Qlesum_{Min mathcal{M}}2=2|mathcal{M}|=2cdot n!.
$$On the other hand, we have
$$
Q=sum_{Ain N}sum_{Min mathcal{M}}1_{Ain M}=sum_{Ain N}|A|!(n-|A|)!
$$ since the number of maximal chains containing $A$ is $|A|!(n-|A|)!$. Note that $j!(n-j)!$ is minimized when $j=lfloor frac{n}{2}rfloor$. Thus, we have
$$
Q= sum_{Ain N}|A|!(n-|A|)!ge lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|
$$ Gathering them together, we get
$$
lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!|N|le 2cdot n!
$$or
$$
|N|le 2cdot frac{n!}{lfloor frac{n}{2}rfloor!left(n-lfloor frac{n}{2}rfloorright)!}=2cdot binom{n}{lfloor frac{n}{2}rfloor},
$$ as desired.
answered Jan 12 at 18:09
SongSong
18.6k21651
edited Jan 12 at 18:50
answered Jan 12 at 18:09
SongSong
18.6k21651
answered Jan 12 at 18:09
SongSong
18.6k21651
answered Jan 12 at 18:09
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
$begingroup$
If $X$ does not contain a $3$-elament chain $Asubset Bsubset C$, then $X$ is the union of two antichains. Namely, the set of all minimal elements of $X$ is an antichain, and the set of all non-minimal elements of $X$ is another antichain. By Sperner's theorem, an antichain of subsets of ${1,2,3,dots,n}$ has at most $binom n{lfloor n/2rfloor}$ elements, so your set $X$ has at most twice that many elements.
answered Jan 13 at 3:16
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
If $X$ does not contain a $3$-elament chain $Asubset Bsubset C$, then $X$ is the union of two antichains. Namely, the set of all minimal elements of $X$ is an antichain, and the set of all non-minimal elements of $X$ is another antichain. By Sperner's theorem, an antichain of subsets of ${1,2,3,dots,n}$ has at most $binom n{lfloor n/2rfloor}$ elements, so your set $X$ has at most twice that many elements.
answered Jan 13 at 3:16
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
If $X$ does not contain a $3$-elament chain $Asubset Bsubset C$, then $X$ is the union of two antichains. Namely, the set of all minimal elements of $X$ is an antichain, and the set of all non-minimal elements of $X$ is another antichain. By Sperner's theorem, an antichain of subsets of ${1,2,3,dots,n}$ has at most $binom n{lfloor n/2rfloor}$ elements, so your set $X$ has at most twice that many elements.
answered Jan 13 at 3:16
bofbof
52.6k559121
$endgroup$
If $X$ does not contain a $3$-elament chain $Asubset Bsubset C$, then $X$ is the union of two antichains. Namely, the set of all minimal elements of $X$ is an antichain, and the set of all non-minimal elements of $X$ is another antichain. By Sperner's theorem, an antichain of subsets of ${1,2,3,dots,n}$ has at most $binom n{lfloor n/2rfloor}$ elements, so your set $X$ has at most twice that many elements.
answered Jan 13 at 3:16
bofbof
52.6k559121
answered Jan 13 at 3:16
bofbof
52.6k559121
answered Jan 13 at 3:16
bofbof
52.6k559121
answered Jan 13 at 3:16
bofbof
52.6k559121
52.6k559121
add a comment |
add a comment |
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