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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, $Filters(L)$ 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 $Filters(L)$ is a consequence of passing to filters and not an easy byproduct of the order on $L$. 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 $f_*$ 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.