Characterization of zero-dimensional frames via lattices of ideals
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My question concerns the left-to-right implication of the following:
Theorem A frame $L$ is compact and zero-dimensional iff it is isomorphic to the lattice of ideals $mathcal{I}(B)$ of some Boolean algebra $B$.
A frame $L$ is a complete and bounded lattice which satisfies: $xwedge bigvee_{iin I}a_i=bigvee_{iin I} xwedge s_i$ (it is a complete Heyting algebra). $ain L$ is complemented iff there is a unique $a^ast$ such that $awedge a^ast=0$ and $avee a^ast=1$. We take $Z(L)$ to be the set of all complemented elements of $L$. $Z(L)$ is a Boolean algebra, and is usually called the center of $L$.
$L$ is zero-dimensional iff for every $ain L$ there exists $Ssubseteq Z(L)$ such that $a=bigvee S$. Finally $L$ is compact iff it's compact as a lattice.
So, in a proof of the theorem we take the lattice of ideals of the center of $L$ and define $hcolon Lrightarrow mathcal{I}(Z(L))$ as $h(a)=(a]cap Z(L)$, where $(a]={xin Lmid xleqslant a}$. Of course, $h(a)$ must be an ideal, but I fail to see how to prove that it is a downward set (the other two properties of ideals are rather obvious: $0in h(a)$ and $h(a)$ must closed for finite joins since $Z(L)$ is a sublattice of $L$). But taking $xin h(a)$ and $yleqslant x$, I cannot see how to prove that $yin Z(L)$.
Could you please help me with this?
ideals boolean-algebra lattice-orders heyting-algebra
asked Nov 28 at 10:31
Mad Hatter
481213
add a comment |
up vote
0
down vote
favorite
My question concerns the left-to-right implication of the following:
Theorem A frame $L$ is compact and zero-dimensional iff it is isomorphic to the lattice of ideals $mathcal{I}(B)$ of some Boolean algebra $B$.
A frame $L$ is a complete and bounded lattice which satisfies: $xwedge bigvee_{iin I}a_i=bigvee_{iin I} xwedge s_i$ (it is a complete Heyting algebra). $ain L$ is complemented iff there is a unique $a^ast$ such that $awedge a^ast=0$ and $avee a^ast=1$. We take $Z(L)$ to be the set of all complemented elements of $L$. $Z(L)$ is a Boolean algebra, and is usually called the center of $L$.
$L$ is zero-dimensional iff for every $ain L$ there exists $Ssubseteq Z(L)$ such that $a=bigvee S$. Finally $L$ is compact iff it's compact as a lattice.
So, in a proof of the theorem we take the lattice of ideals of the center of $L$ and define $hcolon Lrightarrow mathcal{I}(Z(L))$ as $h(a)=(a]cap Z(L)$, where $(a]={xin Lmid xleqslant a}$. Of course, $h(a)$ must be an ideal, but I fail to see how to prove that it is a downward set (the other two properties of ideals are rather obvious: $0in h(a)$ and $h(a)$ must closed for finite joins since $Z(L)$ is a sublattice of $L$). But taking $xin h(a)$ and $yleqslant x$, I cannot see how to prove that $yin Z(L)$.
Could you please help me with this?
ideals boolean-algebra lattice-orders heyting-algebra
asked Nov 28 at 10:31
Mad Hatter
481213
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question concerns the left-to-right implication of the following:
Theorem A frame $L$ is compact and zero-dimensional iff it is isomorphic to the lattice of ideals $mathcal{I}(B)$ of some Boolean algebra $B$.
A frame $L$ is a complete and bounded lattice which satisfies: $xwedge bigvee_{iin I}a_i=bigvee_{iin I} xwedge s_i$ (it is a complete Heyting algebra). $ain L$ is complemented iff there is a unique $a^ast$ such that $awedge a^ast=0$ and $avee a^ast=1$. We take $Z(L)$ to be the set of all complemented elements of $L$. $Z(L)$ is a Boolean algebra, and is usually called the center of $L$.
$L$ is zero-dimensional iff for every $ain L$ there exists $Ssubseteq Z(L)$ such that $a=bigvee S$. Finally $L$ is compact iff it's compact as a lattice.
So, in a proof of the theorem we take the lattice of ideals of the center of $L$ and define $hcolon Lrightarrow mathcal{I}(Z(L))$ as $h(a)=(a]cap Z(L)$, where $(a]={xin Lmid xleqslant a}$. Of course, $h(a)$ must be an ideal, but I fail to see how to prove that it is a downward set (the other two properties of ideals are rather obvious: $0in h(a)$ and $h(a)$ must closed for finite joins since $Z(L)$ is a sublattice of $L$). But taking $xin h(a)$ and $yleqslant x$, I cannot see how to prove that $yin Z(L)$.
Could you please help me with this?
ideals boolean-algebra lattice-orders heyting-algebra
asked Nov 28 at 10:31
Mad Hatter
481213
My question concerns the left-to-right implication of the following:
Theorem A frame $L$ is compact and zero-dimensional iff it is isomorphic to the lattice of ideals $mathcal{I}(B)$ of some Boolean algebra $B$.
A frame $L$ is a complete and bounded lattice which satisfies: $xwedge bigvee_{iin I}a_i=bigvee_{iin I} xwedge s_i$ (it is a complete Heyting algebra). $ain L$ is complemented iff there is a unique $a^ast$ such that $awedge a^ast=0$ and $avee a^ast=1$. We take $Z(L)$ to be the set of all complemented elements of $L$. $Z(L)$ is a Boolean algebra, and is usually called the center of $L$.
$L$ is zero-dimensional iff for every $ain L$ there exists $Ssubseteq Z(L)$ such that $a=bigvee S$. Finally $L$ is compact iff it's compact as a lattice.
So, in a proof of the theorem we take the lattice of ideals of the center of $L$ and define $hcolon Lrightarrow mathcal{I}(Z(L))$ as $h(a)=(a]cap Z(L)$, where $(a]={xin Lmid xleqslant a}$. Of course, $h(a)$ must be an ideal, but I fail to see how to prove that it is a downward set (the other two properties of ideals are rather obvious: $0in h(a)$ and $h(a)$ must closed for finite joins since $Z(L)$ is a sublattice of $L$). But taking $xin h(a)$ and $yleqslant x$, I cannot see how to prove that $yin Z(L)$.
Could you please help me with this?
ideals boolean-algebra lattice-orders heyting-algebra
ideals boolean-algebra lattice-orders heyting-algebra
asked Nov 28 at 10:31
Mad Hatter
481213
asked Nov 28 at 10:31
Mad Hatter
481213
asked Nov 28 at 10:31
Mad Hatter
481213
asked Nov 28 at 10:31
Mad Hatter
481213
asked Nov 28 at 10:31
Mad Hatter
481213
481213
add a comment |
add a comment |
1 Answer
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You want to prove that $h(a)$ is an ideal of $Z(L)$. Thus, it has to be downward closed in $Z(L)$, not in $L$.
Showing that $h(a)$ is downward closed in $Z(L)$ is very easy because $h(a)$ is the intersection of a downward closed subset of $L$ with $Z(L)$.
answered Dec 12 at 22:34
Luca Carai
1948
Blind me ;). Thanks!
– Mad Hatter
2 days ago
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You want to prove that $h(a)$ is an ideal of $Z(L)$. Thus, it has to be downward closed in $Z(L)$, not in $L$.
Showing that $h(a)$ is downward closed in $Z(L)$ is very easy because $h(a)$ is the intersection of a downward closed subset of $L$ with $Z(L)$.
answered Dec 12 at 22:34
Luca Carai
1948
Blind me ;). Thanks!
– Mad Hatter
2 days ago
add a comment |
up vote
1
down vote
accepted
You want to prove that $h(a)$ is an ideal of $Z(L)$. Thus, it has to be downward closed in $Z(L)$, not in $L$.
Showing that $h(a)$ is downward closed in $Z(L)$ is very easy because $h(a)$ is the intersection of a downward closed subset of $L$ with $Z(L)$.
answered Dec 12 at 22:34
Luca Carai
1948
Blind me ;). Thanks!
– Mad Hatter
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You want to prove that $h(a)$ is an ideal of $Z(L)$. Thus, it has to be downward closed in $Z(L)$, not in $L$.
Showing that $h(a)$ is downward closed in $Z(L)$ is very easy because $h(a)$ is the intersection of a downward closed subset of $L$ with $Z(L)$.
answered Dec 12 at 22:34
Luca Carai
1948
You want to prove that $h(a)$ is an ideal of $Z(L)$. Thus, it has to be downward closed in $Z(L)$, not in $L$.
Showing that $h(a)$ is downward closed in $Z(L)$ is very easy because $h(a)$ is the intersection of a downward closed subset of $L$ with $Z(L)$.
answered Dec 12 at 22:34
Luca Carai
1948
answered Dec 12 at 22:34
Luca Carai
1948
answered Dec 12 at 22:34
Luca Carai
1948
answered Dec 12 at 22:34
Luca Carai
1948
1948
Blind me ;). Thanks!
– Mad Hatter
2 days ago
add a comment |
Blind me ;). Thanks!
– Mad Hatter
2 days ago
Blind me ;). Thanks!
– Mad Hatter
2 days ago
Blind me ;). Thanks!
– Mad Hatter
2 days ago
add a comment |
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