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1. • CommentRowNumber1.
   • CommentAuthorJohn Baez
   • CommentTimeJun 7th 2015
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   Author: John Baez
   Format: MarkdownItexI added a description of the opposite of the category of finite distributive lattices to [[distributive lattice]].

   I added a description of the opposite of the category of finite distributive lattices to distributive lattice.

2. • CommentRowNumber2.
   • CommentAuthorporton
   • CommentTimeOct 17th 2018
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   Author: porton
   Format: MarkdownItexDoes it make sense to define “infinitely distributive property” for non-complete lattices? (Something like: “Whenever the join exists, it satisfies the infinite distributive law.”) Does this work in practice? (is a useful definition?)

   Does it make sense to define “infinitely distributive property” for non-complete lattices? (Something like: “Whenever the join exists, it satisfies the infinite distributive law.”)

   Does this work in practice? (is a useful definition?)

3. • CommentRowNumber3.
   • CommentAuthorOscar_Cunningham
   • CommentTimeOct 17th 2018
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   Author: Oscar_Cunningham
   Format: MarkdownItexIn categories it's often better to say "has a left adjoint" than "preserves small limits" in the case when the domain category doesn't have all small limits. So the infinite distributive property for a noncomplete lattice might be most nicely defined as "for all $x$, the function $y\mapsto x\wedge y$ has a left adjoint". This is precisely the definition of a [[Heyting algebra]].

   In categories it’s often better to say “has a left adjoint” than “preserves small limits” in the case when the domain category doesn’t have all small limits. So the infinite distributive property for a noncomplete lattice might be most nicely defined as “for all $x$, the function $y\mapsto x\wedge y$ has a left adjoint”. This is precisely the definition of a Heyting algebra.

4. • CommentRowNumber4.
   • CommentAuthorporton
   • CommentTimeOct 17th 2018
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   Author: porton
   Format: MarkdownItex@Oscar_Cunningham How to prove that distributivity over all existing joins implies that it's adjoint? As far as I know, distributivity over all existing joins does not imply adjointness. (However, it implies adjointness in the case of a complete lattice.)

   @Oscar_Cunningham

   How to prove that distributivity over all existing joins implies that it’s adjoint?

   As far as I know, distributivity over all existing joins does not imply adjointness. (However, it implies adjointness in the case of a complete lattice.)

5. • CommentRowNumber5.
   • CommentAuthorOscar_Cunningham
   • CommentTimeOct 17th 2018
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   Author: Oscar_Cunningham
   Format: MarkdownItexDistributivity over all existing joins doesn't imply adjointness, I'm just saying that adjointness is a better behaved notion than distributivity over joins, and might be more useful in practice.

   Distributivity over all existing joins doesn’t imply adjointness, I’m just saying that adjointness is a better behaved notion than distributivity over joins, and might be more useful in practice.