Mixing
Maximum cardinality of a set of subsets
$begingroup$
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
$endgroup$
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
$endgroup$
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
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
12.5k631
$endgroup$
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
51.7k558120
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071187%2fmaximum-cardinality-of-a-set-of-subsets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
12.5k631
$endgroup$
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
12.5k631
$endgroup$
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
12.5k631
$endgroup$
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
12.5k631
edited Jan 12 at 18:50
answered Jan 12 at 18:09
SongSong
12.5k631
answered Jan 12 at 18:09
SongSong
12.5k631
answered Jan 12 at 18:09
SongSong
12.5k631
12.5k631
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
51.7k558120
$endgroup$
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
51.7k558120
$endgroup$
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
51.7k558120
$endgroup$
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
51.7k558120
answered Jan 13 at 3:16
bofbof
51.7k558120
answered Jan 13 at 3:16
bofbof
51.7k558120
answered Jan 13 at 3:16
bofbof
51.7k558120
51.7k558120
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071187%2fmaximum-cardinality-of-a-set-of-subsets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
