Representation theorem for Heyting algebras?
A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).
I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?
Thank you in advance.
reference-request logic
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
add a comment |
A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).
I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?
Thank you in advance.
reference-request logic
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33
The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57
Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22
1
Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24
add a comment |
A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).
I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?
Thank you in advance.
reference-request logic
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).
I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?
Thank you in advance.
reference-request logic
reference-request logic
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
edited Jul 14 '14 at 17:11
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
asked Jul 14 '14 at 16:55
Marco Vergura
3,0811930
3,0811930
@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33
The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57
Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22
1
Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24
add a comment |
@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33
The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57
Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22
1
Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24
@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33
@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33
The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57
The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57
Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22
Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22
1
1
Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24
Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24
add a comment |
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Esakia Duality might just be the thing that you are looking for.
For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
leq,
mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.
This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.
Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.
answered Feb 14 '15 at 12:10
FML
18618
add a comment |
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Esakia Duality might just be the thing that you are looking for.
For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
leq,
mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.
This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.
Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.
answered Feb 14 '15 at 12:10
FML
18618
add a comment |
Esakia Duality might just be the thing that you are looking for.
For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
leq,
mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.
This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.
Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.
answered Feb 14 '15 at 12:10
FML
18618
add a comment |
Esakia Duality might just be the thing that you are looking for.
For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
leq,
mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.
This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.
Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.
answered Feb 14 '15 at 12:10
FML
18618
Esakia Duality might just be the thing that you are looking for.
For every Heyting algebra $A$ there exists a so called Esakia space $mathscr{X}=(X,
leq,
mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $mathscr{X}$.
This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.
Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.
answered Feb 14 '15 at 12:10
FML
18618
answered Feb 14 '15 at 12:10
FML
18618
answered Feb 14 '15 at 12:10
FML
18618
answered Feb 14 '15 at 12:10
FML
18618
18618
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@MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras.
– Kyle
Jul 14 '14 at 17:33
The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points".
– Zhen Lin
Jul 14 '14 at 17:57
Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for?
– Nagase
Jul 25 '14 at 5:22
1
Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = mathfrak S(B)$ [McKinsey and Tarski, 1946]" ?
– Mauro ALLEGRANZA
Aug 29 '14 at 14:24