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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 29th 2013
   • (edited Jun 29th 2013)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexIn the article [[compactum]], under Stone-Cech compactification, it is stated that the ultrafilter monad is commutative. I really don't think that's true (and it has nothing to do with $\beta S$ being a compactum). To say $\beta$ is commutative is more or less saying that all the definable operations $X^J \to X$ (parametrized by elements $\theta$ of the set $\beta J$) on a $\beta$-algebra $X$ are continuous. Explicitly, the associated operation is given by the composite $$X^J = \hom(J, X) \to \hom(\beta J, \beta X) \stackrel{\hom(\theta, 1)}{\to} \hom(1, \beta X) \cong \beta X \stackrel{\alpha}{\to} X$$ where $\alpha: \beta X \to X$ is the algebra structure. I am reasonably confident all these operations are distinct even for a simple case like $X = \mathbf{2}$, the two-point discrete space (**edit:** indeed, it may be shown that the operation $\mathbf{2}^J \to \mathbf{2}$ coincides with the ultrafilter $\theta$ viewed as Boolean algebra map $\mathbf{2}^J \to \mathbf{2}$ -- details available for anyone interested). For example, if $J = \mathbb{N}$, then there would be $\beta \mathbb{N}$ many such operations, which has cardinality $2^c$. On the other hand, there are only countably many continuous maps $\mathbf{2}^{\mathbb{N}} \to \mathbf{2}$. This is because $\mathbf{2}^{\mathbb{N}}$ is the Stone space attached to the free Boolean algebra $Bool(\mathbb{N})$ on countably many elements, and we retrieve a Boolean algebra $B$ from its Stone space $Stone(B)$ by taking the set of continuous maps $Stone(B) \to \mathbf{2}$, equipped with the pointwise Boolean algebra structure. In other words, the hom-set $Top(\mathbf{2}^{\mathbb{N}}, \mathbf{2})$ is in natural bijection with the countable Boolean algebra $Bool(\mathbb{N})$. If there are no objections, I'll go in and fix this.

   In the article compactum, under Stone-Cech compactification, it is stated that the ultrafilter monad is commutative. I really don’t think that’s true (and it has nothing to do with being a compactum).

   To say is commutative is more or less saying that all the definable operations (parametrized by elements of the set ) on a -algebra are continuous. Explicitly, the associated operation is given by the composite

   where is the algebra structure. I am reasonably confident all these operations are distinct even for a simple case like , the two-point discrete space (edit: indeed, it may be shown that the operation coincides with the ultrafilter viewed as Boolean algebra map – details available for anyone interested). For example, if , then there would be many such operations, which has cardinality .

   On the other hand, there are only countably many continuous maps . This is because is the Stone space attached to the free Boolean algebra on countably many elements, and we retrieve a Boolean algebra from its Stone space by taking the set of continuous maps , equipped with the pointwise Boolean algebra structure. In other words, the hom-set is in natural bijection with the countable Boolean algebra .

   If there are no objections, I’ll go in and fix this.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 29th 2013
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   Author: Todd_Trimble
   Format: MarkdownItexI have now made some edits to the section on Stone-Cech compactification, and I made a new section on the category of compact Hausdorff spaces and its wonderful properties.

   I have now made some edits to the section on Stone-Cech compactification, and I made a new section on the category of compact Hausdorff spaces and its wonderful properties.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeJun 30th 2013
   • (edited Jun 30th 2013)
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   Author: Mike Shulman
   Format: MarkdownItexGiraud's theorem also requires infinite coproducts to be disjoint and stable. Are they?

   Giraud’s theorem also requires infinite coproducts to be disjoint and stable. Are they?

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 30th 2013
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   Author: Todd_Trimble
   Format: MarkdownItexOh, right (thanks). Oops. Well, I doubt it.

   Oh, right (thanks). Oops. Well, I doubt it.

5. • CommentRowNumber5.
   • CommentAuthorZhen Lin
   • CommentTimeJun 30th 2013
   • PermaLink
   Author: Zhen Lin
   Format: MarkdownItexIf we follow the conventions in the Elephant, a pretopos only needs to be finitary-extensive.

   If we follow the conventions in the Elephant, a pretopos only needs to be finitary-extensive.

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 30th 2013
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   Author: Todd_Trimble
   Format: MarkdownItex@Zhen: yes, but I think Mike might have been alluding to my rash comment on Giraud's theorem implying that the only obstacle to $Comp$ (or $CH$) being a Grothendieck topos is that it doesn't have a small generating family. And I think I can turn my doubt in #4 to a positive "no". If infinite coproducts were stable under pullbacks, then (for example) we'd have that $X \times -: CH \to CH$ preserves colimits (right?). But since $CH$ is both total and cototal, this would be enough to conclude $X \times -$ has a right adjoint (I think it's totality that's relevanthere, but never mind). But $CH$ is not cartesian closed.

   @Zhen: yes, but I think Mike might have been alluding to my rash comment on Giraud’s theorem implying that the only obstacle to (or ) being a Grothendieck topos is that it doesn’t have a small generating family.

   And I think I can turn my doubt in #4 to a positive “no”. If infinite coproducts were stable under pullbacks, then (for example) we’d have that preserves colimits (right?). But since is both total and cototal, this would be enough to conclude has a right adjoint (I think it’s totality that’s relevanthere, but never mind). But is not cartesian closed.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeJun 30th 2013
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexRight. A pretopos is finitely extensive, but the hypothesis of Giraud's theorem is that the category is an infinitary-pretopos, which must by definition be infinitary-extensive.

   Right. A pretopos is finitely extensive, but the hypothesis of Giraud’s theorem is that the category is an infinitary-pretopos, which must by definition be infinitary-extensive.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeMay 3rd 2017
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have edited the Idea-section at _[[compactum]]_ slightly, for readability. Since "[[compact Hausdorff space]]" redirects to this entry, I think the Idea-section should be clear about this right away. The Idea section also has, and used to have, the following lines in it: > One may consider the analogous condition for [[convergence spaces]], or for [[locales]]. Even though these are all different contexts, the resulting notion of compactum is (at least assuming the [[axiom of choice]]) always the same. What is actually meant by this ("always the same")? The entry never comes back to this statement. I suppose if a topological space is compact Hausdorff as a topological space, then it is also compact Hausdorff when regarded as a locale (?). But the claim is not that all compact Hausdorff locales are spatial locales coming from compact Hausdorff spaces, or is it?

   I have edited the Idea-section at compactum slightly, for readability. Since “compact Hausdorff space” redirects to this entry, I think the Idea-section should be clear about this right away.

   The Idea section also has, and used to have, the following lines in it:

   One may consider the analogous condition for convergence spaces, or for locales. Even though these are all different contexts, the resulting notion of compactum is (at least assuming the axiom of choice) always the same.

   What is actually meant by this (“always the same”)? The entry never comes back to this statement. I suppose if a topological space is compact Hausdorff as a topological space, then it is also compact Hausdorff when regarded as a locale (?). But the claim is not that all compact Hausdorff locales are spatial locales coming from compact Hausdorff spaces, or is it?

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeMay 3rd 2017
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   Author: Urs
   Format: MarkdownItex> But the claim is not that all compact Hausdorff locales are spatial locales coming from compact Hausdorff spaces, or is it? Oh, I see from _[[locally compact locale]]_ that it is.

   But the claim is not that all compact Hausdorff locales are spatial locales coming from compact Hausdorff spaces, or is it?

   Oh, I see from locally compact locale that it is.