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1. • CommentRowNumber1.
   • CommentAuthortyler bryson
   • CommentTimeAug 9th 2016
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   Author: tyler bryson
   Format: TextThe present nLab page for Filter (https://ncatlab.org/nlab/show/filter) claims the following: "If L is a complete join-semilattice, then Filters(L) is a complete lattice." I don't see how this is true and no reference is given. Can someone shed light on this? After two days with it, I suspect this is a typo...but am not sure. Much obliged, Tyler Bryson

   The present nLab page for Filter (https://ncatlab.org/nlab/show/filter) claims the following:
   
   "If L is a complete join-semilattice, then Filters(L) is a complete lattice."
   
   I don't see how this is true and no reference is given. Can someone shed light on this? After two days with it, I suspect this is a typo...but am not sure.
   
   Much obliged,
   Tyler Bryson
2. • CommentRowNumber2.
   • CommentAuthorDexter Chua
   • CommentTimeAug 9th 2016
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   Author: Dexter Chua
   Format: MarkdownItexA complete join-semilattice is automatically a complete meet-semilattice, so by the first point in the list, $Filters(L)$ is a complete join-semillatice, hence a complete lattice.

   A complete join-semilattice is automatically a complete meet-semilattice, so by the first point in the list, is a complete join-semillatice, hence a complete lattice.

3. • CommentRowNumber3.
   • CommentAuthortyler bryson
   • CommentTimeAug 9th 2016
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   Author: tyler bryson
   Format: MarkdownItexThank you. I should have seen that. I do think the wording on that bullet point could be improved. As it stands, I expect I am not the only person who'll incorrectly infer that the completeness of the lattice $Filters(L)$ is a consequence of passing to filters and not an easy byproduct of the order on $L$. I may submit an edit.

   Thank you. I should have seen that. I do think the wording on that bullet point could be improved. As it stands, I expect I am not the only person who’ll incorrectly infer that the completeness of the lattice is a consequence of passing to filters and not an easy byproduct of the order on . I may submit an edit.

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeAug 27th 2016
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   Author: TobyBartels
   Format: MarkdownItexI think that the point of those bullet points is supposed to be about how $f_*$ has various nice properties when $f$ does. So the first sentence of each is sort of just introductory: if $L$ has some structure, then $Filters(L)$ has some structure, OK, but now here\'s what\'s interesting: if $f$ preserves the structure on $L$, then $f_*$ preserves the structure on $Filters(L)$, and *that\'s* what we want to talk about! At least, that\'s the attitude, I think.

   I think that the point of those bullet points is supposed to be about how has various nice properties when does. So the first sentence of each is sort of just introductory: if has some structure, then has some structure, OK, but now here's what's interesting: if preserves the structure on , then preserves the structure on , and that's what we want to talk about! At least, that's the attitude, I think.