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Covering relation in the lattice of closed subsets
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Covering relation in the lattice of closed subsets
Hello!
I am currently reading Martin Aigner's Combinatorial Theory and I have to say this book is definitely above my level, but (very) slowly things are coming together. Anyway I got to one statement that doesn't seem obvious to me. There is a proof given, but I don't see how it is sufficient. The statement is the following:
$overlineAcup p cdot!! >A$ for any $Ain L(S)$, $pnotin A$.
Here we are talking about a matroid on the set $S$. $overlineA$ is the closure of $A$ and $L(S)$ denotes the lattice of closed subsets of $S$. We use the symbol $cdot!! >$ for the covering relation.
The proof given is the following:
If $qnotin A$, $qin overlineAcup p$, then by the exchange property of the closure operator $pin overlineAcup q$, i. e. $overlineAcup p = overlineAcup q$.
That's it. Now I have this question (among many): Isn't it possible for example to consider another element $r$, such that $rnotin A$, $rin overlineAcup p$ and $rnotin overlineAcup q$? Now $overlineAcup p ne overlineAcup q$.
I fear that I am missing something really obvious so I decided to ask.
order-theory lattice-orders matroids
asked Jul 19 at 12:32
batboio
745
add a comment |Â
up vote
1
down vote
favorite
Covering relation in the lattice of closed subsets
Hello!
I am currently reading Martin Aigner's Combinatorial Theory and I have to say this book is definitely above my level, but (very) slowly things are coming together. Anyway I got to one statement that doesn't seem obvious to me. There is a proof given, but I don't see how it is sufficient. The statement is the following:
$overlineAcup p cdot!! >A$ for any $Ain L(S)$, $pnotin A$.
Here we are talking about a matroid on the set $S$. $overlineA$ is the closure of $A$ and $L(S)$ denotes the lattice of closed subsets of $S$. We use the symbol $cdot!! >$ for the covering relation.
The proof given is the following:
If $qnotin A$, $qin overlineAcup p$, then by the exchange property of the closure operator $pin overlineAcup q$, i. e. $overlineAcup p = overlineAcup q$.
That's it. Now I have this question (among many): Isn't it possible for example to consider another element $r$, such that $rnotin A$, $rin overlineAcup p$ and $rnotin overlineAcup q$? Now $overlineAcup p ne overlineAcup q$.
I fear that I am missing something really obvious so I decided to ask.
order-theory lattice-orders matroids
asked Jul 19 at 12:32
batboio
745
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Covering relation in the lattice of closed subsets
Hello!
I am currently reading Martin Aigner's Combinatorial Theory and I have to say this book is definitely above my level, but (very) slowly things are coming together. Anyway I got to one statement that doesn't seem obvious to me. There is a proof given, but I don't see how it is sufficient. The statement is the following:
$overlineAcup p cdot!! >A$ for any $Ain L(S)$, $pnotin A$.
Here we are talking about a matroid on the set $S$. $overlineA$ is the closure of $A$ and $L(S)$ denotes the lattice of closed subsets of $S$. We use the symbol $cdot!! >$ for the covering relation.
The proof given is the following:
If $qnotin A$, $qin overlineAcup p$, then by the exchange property of the closure operator $pin overlineAcup q$, i. e. $overlineAcup p = overlineAcup q$.
That's it. Now I have this question (among many): Isn't it possible for example to consider another element $r$, such that $rnotin A$, $rin overlineAcup p$ and $rnotin overlineAcup q$? Now $overlineAcup p ne overlineAcup q$.
I fear that I am missing something really obvious so I decided to ask.
order-theory lattice-orders matroids
asked Jul 19 at 12:32
batboio
745
Covering relation in the lattice of closed subsets
Hello!
I am currently reading Martin Aigner's Combinatorial Theory and I have to say this book is definitely above my level, but (very) slowly things are coming together. Anyway I got to one statement that doesn't seem obvious to me. There is a proof given, but I don't see how it is sufficient. The statement is the following:
$overlineAcup p cdot!! >A$ for any $Ain L(S)$, $pnotin A$.
Here we are talking about a matroid on the set $S$. $overlineA$ is the closure of $A$ and $L(S)$ denotes the lattice of closed subsets of $S$. We use the symbol $cdot!! >$ for the covering relation.
The proof given is the following:
If $qnotin A$, $qin overlineAcup p$, then by the exchange property of the closure operator $pin overlineAcup q$, i. e. $overlineAcup p = overlineAcup q$.
That's it. Now I have this question (among many): Isn't it possible for example to consider another element $r$, such that $rnotin A$, $rin overlineAcup p$ and $rnotin overlineAcup q$? Now $overlineAcup p ne overlineAcup q$.
I fear that I am missing something really obvious so I decided to ask.
order-theory lattice-orders matroids
asked Jul 19 at 12:32
batboio
745
asked Jul 19 at 12:32
batboio
745
asked Jul 19 at 12:32
batboio
745
asked Jul 19 at 12:32
batboio
745
745
add a comment |Â
add a comment |Â
1 Answer
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1
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It might help if you try to write your own proof. The key idea is the same as in the text. I have written this one out in (too much) detail so you can see what I mean.
Suppose $B$ is a closed set such that $Asubseteq B subseteq overlineAcup p$. If there is $qin B$ such that $qnotin A$, then $qinoverlineAcup qsubseteq B subseteq overlineAcup p$. By the exchange property, $pin overlineAcup qsubseteq B$. Since $Asubseteq B$ and $pin B$, we get that $overlineAcup psubseteq B$, hence $overlineAcup p=B$. This shows that if $B$ is between $A$ and $overlineAcup p$ and $Bneq A$, then $B=overlineAcup p$, i.e. that $overlineAcup p$ covers $A$ in the lattice of closed sets.
answered Jul 19 at 23:17
Eran
1,166818
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It might help if you try to write your own proof. The key idea is the same as in the text. I have written this one out in (too much) detail so you can see what I mean.
Suppose $B$ is a closed set such that $Asubseteq B subseteq overlineAcup p$. If there is $qin B$ such that $qnotin A$, then $qinoverlineAcup qsubseteq B subseteq overlineAcup p$. By the exchange property, $pin overlineAcup qsubseteq B$. Since $Asubseteq B$ and $pin B$, we get that $overlineAcup psubseteq B$, hence $overlineAcup p=B$. This shows that if $B$ is between $A$ and $overlineAcup p$ and $Bneq A$, then $B=overlineAcup p$, i.e. that $overlineAcup p$ covers $A$ in the lattice of closed sets.
answered Jul 19 at 23:17
Eran
1,166818
add a comment |Â
up vote
1
down vote
accepted
It might help if you try to write your own proof. The key idea is the same as in the text. I have written this one out in (too much) detail so you can see what I mean.
Suppose $B$ is a closed set such that $Asubseteq B subseteq overlineAcup p$. If there is $qin B$ such that $qnotin A$, then $qinoverlineAcup qsubseteq B subseteq overlineAcup p$. By the exchange property, $pin overlineAcup qsubseteq B$. Since $Asubseteq B$ and $pin B$, we get that $overlineAcup psubseteq B$, hence $overlineAcup p=B$. This shows that if $B$ is between $A$ and $overlineAcup p$ and $Bneq A$, then $B=overlineAcup p$, i.e. that $overlineAcup p$ covers $A$ in the lattice of closed sets.
answered Jul 19 at 23:17
Eran
1,166818
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It might help if you try to write your own proof. The key idea is the same as in the text. I have written this one out in (too much) detail so you can see what I mean.
Suppose $B$ is a closed set such that $Asubseteq B subseteq overlineAcup p$. If there is $qin B$ such that $qnotin A$, then $qinoverlineAcup qsubseteq B subseteq overlineAcup p$. By the exchange property, $pin overlineAcup qsubseteq B$. Since $Asubseteq B$ and $pin B$, we get that $overlineAcup psubseteq B$, hence $overlineAcup p=B$. This shows that if $B$ is between $A$ and $overlineAcup p$ and $Bneq A$, then $B=overlineAcup p$, i.e. that $overlineAcup p$ covers $A$ in the lattice of closed sets.
answered Jul 19 at 23:17
Eran
1,166818
It might help if you try to write your own proof. The key idea is the same as in the text. I have written this one out in (too much) detail so you can see what I mean.
Suppose $B$ is a closed set such that $Asubseteq B subseteq overlineAcup p$. If there is $qin B$ such that $qnotin A$, then $qinoverlineAcup qsubseteq B subseteq overlineAcup p$. By the exchange property, $pin overlineAcup qsubseteq B$. Since $Asubseteq B$ and $pin B$, we get that $overlineAcup psubseteq B$, hence $overlineAcup p=B$. This shows that if $B$ is between $A$ and $overlineAcup p$ and $Bneq A$, then $B=overlineAcup p$, i.e. that $overlineAcup p$ covers $A$ in the lattice of closed sets.
answered Jul 19 at 23:17
Eran
1,166818
answered Jul 19 at 23:17
Eran
1,166818
answered Jul 19 at 23:17
Eran
1,166818
answered Jul 19 at 23:17
Eran
1,166818
1,166818
add a comment |Â
add a comment |Â
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