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Reply | Forward Message #188 of 654 |
I have a problem related to duality theory for Heyting algebras. It is known
that open sets of a topological space form a Heyting algebra under natural
operations. Such an algebra is
always: (1) complete as a lattice, (2) every element can be represented as
an
infimum of some family of meet-prime elements - i.e. elements indecomposable
by finite meets. I don't know if property (2) has
any particular name, for the time being let's call it, say,
quasi-simplicity. On the other hand, given any Heyting algebra A which is
complete and quasi-simple, one may construct a topological space (even a
T_0-space) whose algebra
of open sets is isomorphic to A. As far as I know, that result was proven by
Dowkert and Popert in Proceedings of London Mathematical Society, 1966.
There is a class of topological spaces of particular importance for the
theory of Heyting algebras and extensions of intuitionistic logic: Kripke
frames. These are topological spaces obtained from posets with right-order
topology: i.e., upward closed subsets of a poset. It is known that the
Heyting algebra associated with such a topology is always (1) complete, (2')
strongly compactly generated (Esakia): i.e., every element can be obtained
as a supremum of some family of completely join-prime elements - element
indecomposable by joins, even infinite joins; conjunction of (1) and (2') is
then stronger than the property of being algebraic Heyting algebra. On the
other hand, given a complete and strongly compactly generated Heyting
algebra A, one may always construct a poset whose algebra of upward closed
subsets is isomorphic to A. I don't know who was the first to discover those
results, but I would attribute them to Leo Esakia.
Now, I have a question which may be an easy one, but somehow I am unable to
answer it. For any complete lattice, (2') implies (2), i.e., the property of
being strongly compactly generated obviously implies quasi-simplicity. It
may be proven either via the above mentioned duality results or directly: it
is enough to observe that with every completely-join-prime element a we may
associate a meet-prime element m(a), defined as the supremum of all
completely join-prime elements b s.t. a is not below b. Every element of a
Heyting algebra satisfying (1) and (2') may be then obtained as an infimum
of a family of elements of the form m(a). Anyhow, proof seems to make
essential use of (1), i.e., completeness.
So how about relationship between (2) and (2') when completeness is not
assumed? It seems by no means obvious that being strongly compactly
generated should still imply quasi-simplicity. And yet despite my efforts, I
have not been able to produce any counterexample.
Please let me know if you know the answer or you know about any work related
to the problem. I would be very obliged for any help.
Best,

Tadeusz Litak
School of Information Science
JAIST







Sun Nov 30, 2003 9:48 am

litak@...
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Message #188 of 654 |
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I have a problem related to duality theory for Heyting algebras. It is known that open sets of a topological space form a Heyting algebra under natural ...
LITAK Tadeusz Michal
litak@...
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Nov 30, 2003
9:48 am

... A. (2') does not imply (2) for arbitrary (bounded, distributive) lattices, but B. (2') does imply (2) for arbitrary Heyting algebras. For A, let L be the...
Keith A. Kearnes
k_kearnes
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Nov 30, 2003
8:55 pm

... A complete Heyting algebra L is also known as a locale, or frame (see, eg, P. Johnstone, Stone Spaces, CUP, 1982), and can be regarded as a generalized...
Pedro Resende
pmr@...
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Nov 30, 2003
10:47 pm

I just noticed that the book The Structure of Finite Algebras, David Hobby and and Ralph McKenzie, American Mathematical Society, 1988, 209 pp. is among those...
Keith A. Kearnes
k_kearnes
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Dec 7, 2003
5:23 am
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