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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeDec 29th 2010
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   Author: Urs
   Format: MarkdownItexadded the previously missing proposition (on equivalent characterizations) to [[essentially small (infinity,1)-category]]

   added the previously missing proposition (on equivalent characterizations) to essentially small (infinity,1)-category

2. • CommentRowNumber2.
   • CommentAuthorTobyBartels
   • CommentTimeDec 30th 2010
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   Author: TobyBartels
   Format: MarkdownItexThe absolute definition (without $\kappa$) did not seem correct to me. Certainly it did not generalise the notion of [[small category]], as stated. I made an edit that should fix this, although it might be more clear to just write the definition out twice.

   The absolute definition (without $\kappa$) did not seem correct to me. Certainly it did not generalise the notion of small category, as stated. I made an edit that should fix this, although it might be more clear to just write the definition out twice.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeDec 30th 2010
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   Author: Mike Shulman
   Format: MarkdownItexAt [[small category]] it says: > A small category structure on a category C is an essentially surjective functor from a set (as a discrete category) to C. A category is essentially small iff it has a small category structure; this does not require the axiom of choice. That doesn't seem right... it doesn't say anything about smallness of the set of morphisms. I would fix it but I don't know the goal of this definition; can someone suggest the right fix?

   At small category it says:

   A small category structure on a category C is an essentially surjective functor from a set (as a discrete category) to C. A category is essentially small iff it has a small category structure; this does not require the axiom of choice.

   That doesn’t seem right… it doesn’t say anything about smallness of the set of morphisms. I would fix it but I don’t know the goal of this definition; can someone suggest the right fix?

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeDec 30th 2010
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   Author: TobyBartels
   Format: MarkdownItexI think that we just want to demand ahead of time that $C$ be locally small. If you wanted to, you could define a small category structure on an arbitrary category $C$ to be the property that $C$ is locally small together with such an eso functor. However, if the purpose of this paragraph is to give a choice-free analogue of the paragraph before it (which I think is likely), then this isn't really necessary. I've fixed it.

   I think that we just want to demand ahead of time that $C$ be locally small. If you wanted to, you could define a small category structure on an arbitrary category $C$ to be the property that $C$ is locally small together with such an eso functor. However, if the purpose of this paragraph is to give a choice-free analogue of the paragraph before it (which I think is likely), then this isn’t really necessary.

   I’ve fixed it.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeDec 30th 2010
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   Author: Urs
   Format: MarkdownItex> The absolute definition (without $\kappa$) did not seem correct to me. Hm, maybe I am mixed up. If a set is $\kappa$-small for _some_ $\kappa$, does that not just mean it is small ?

   The absolute definition (without $\kappa$) did not seem correct to me.

   Hm, maybe I am mixed up. If a set is $\kappa$-small for some $\kappa$, does that not just mean it is small ?

6. • CommentRowNumber6.
   • CommentAuthorTobyBartels
   • CommentTimeDec 30th 2010
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   Author: TobyBartels
   Format: MarkdownItexGee, I guess that it *does* mean that; it seemed such an odd way of saying so. But of course it is true. (Somewhere in my memory is a stray negation operator —or something— still telling me that your version was wrong, but that is now clearly an error.) Changed back, and parenthetical remark added.

   Gee, I guess that it does mean that; it seemed such an odd way of saying so. But of course it is true. (Somewhere in my memory is a stray negation operator —or something— still telling me that your version was wrong, but that is now clearly an error.)

   Changed back, and parenthetical remark added.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeDec 30th 2010
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   Author: Mike Shulman
   Format: MarkdownItexAlthough I agree that both are correct, I prefer the other version. The current version depends on the nontrivial fact that there exist arbitrarily large regular cardinals (or regular sets-of-cardinals), which (in particular) I don't know how to prove constructively without essentially assuming it as an axiom.

   Although I agree that both are correct, I prefer the other version. The current version depends on the nontrivial fact that there exist arbitrarily large regular cardinals (or regular sets-of-cardinals), which (in particular) I don’t know how to prove constructively without essentially assuming it as an axiom.

8. • CommentRowNumber8.
   • CommentAuthorTobyBartels
   • CommentTimeJan 2nd 2011
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   Author: TobyBartels
   Format: Markdown@ Mike What is a reference (or a summary of the argument) for the result (with choice) that there are enough regular cardinals?

   @ Mike

   What is a reference (or a summary of the argument) for the result (with choice) that there are enough regular cardinals?

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeJan 2nd 2011
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   Author: Mike Shulman
   Format: MarkdownItexIf I may quote from [[regular cardinal]]: > The [[successor]] of any infinite cardinal, such as $\aleph_1$, is a regular cardinal. (This requires the [[axiom of choice]].) ... Note that this implies that there exist arbitarily large regular cardinals: for any cardinal $\lambda$ there is a greater regular cardinal, namely $\lambda^+$. Is that enough, or did you want the argument that successors are regular?

   If I may quote from regular cardinal:

   The successor of any infinite cardinal, such as $\aleph_1$, is a regular cardinal. (This requires the axiom of choice.) … Note that this implies that there exist arbitarily large regular cardinals: for any cardinal $\lambda$ there is a greater regular cardinal, namely $\lambda^+$.

   Is that enough, or did you want the argument that successors are regular?

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeJan 3rd 2011
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    Author: TobyBartels
    Format: MarkdownItexOK, thanks, I should have known that.

    OK, thanks, I should have known that.

11. • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeJan 8th 2011
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    Author: TobyBartels
    Format: MarkdownItex>The current version depends on the nontrivial fact that there exist arbitrarily large regular cardinals (or regular sets-of-cardinals), which (in particular) I don't know how to prove constructively without essentially assuming it as an axiom. This axiom is right up top in the cited section (page 10-1) of [the reference by Aczel & Rathjen](http://www.ml.kva.se/preprints/archive/2000-2001/2000-2001-40.pdf) cited by Daniel Méhkeri in his answer to [my M.O. question on large cardinals](http://mathoverflow.net/questions/50143/cardinal-numbers-vs-collections-of-cardinal-numbers).

    The current version depends on the nontrivial fact that there exist arbitrarily large regular cardinals (or regular sets-of-cardinals), which (in particular) I don’t know how to prove constructively without essentially assuming it as an axiom.

    This axiom is right up top in the cited section (page 10-1) of the reference by Aczel & Rathjen cited by Daniel Méhkeri in his answer to my M.O. question on large cardinals.

12. • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeJan 10th 2011
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    Author: Mike Shulman
    Format: MarkdownItexI mentioned [[regular extension axiom]] at [[regular cardinal]] and made it a redirect, and edited [[essentially small (∞,1)-category]] to mention both characterizations. But feel free to object.

    I mentioned regular extension axiom at regular cardinal and made it a redirect, and edited essentially small (∞,1)-category to mention both characterizations. But feel free to object.

13. • CommentRowNumber13.
    • CommentAuthorTobyBartels
    • CommentTimeJan 11th 2011
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    Author: TobyBartels
    Format: MarkdownItexDavid suggested a correction at [[regular cardinal]], which I have approved. Now I have a question there.

    David suggested a correction at regular cardinal, which I have approved. Now I have a question there.

14. • CommentRowNumber14.
    • CommentAuthorMike Shulman
    • CommentTimeJan 11th 2011
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    Author: Mike Shulman
    Format: MarkdownItexInteresting question! Suppose $\kappa$ is a limit cardinal which is definable, i.e there is a first-order formula $\phi$ such that $\exists! \kappa. \phi(\kappa)$ is provable in ZFC. If there are no weakly inaccessible cardinals, then $\kappa$ must be singular. If there are weakly inaccessible cardinals, let $\lambda$ be the smallest one; then $L_\lambda$ is a model of "ZFC + there are no weakly inacessible cardinals" and thus $\kappa^{L_\lambda}$ (the unique element of $L_\lambda$ satisfying $\phi$) is singular in $L_\lambda$. But I guess it's not clear how the singularity of $\kappa^{L_\lambda}$ in $L_\lambda$ might be related to singularity of $\kappa$ in $V$. Maybe the set-theorists on MO would have a thought.

    Interesting question! Suppose $\kappa$ is a limit cardinal which is definable, i.e there is a first-order formula $\phi$ such that $\exists! \kappa. \phi(\kappa)$ is provable in ZFC. If there are no weakly inaccessible cardinals, then $\kappa$ must be singular. If there are weakly inaccessible cardinals, let $\lambda$ be the smallest one; then $L_\lambda$ is a model of “ZFC + there are no weakly inacessible cardinals” and thus $\kappa^{L_\lambda}$ (the unique element of $L_\lambda$ satisfying $\phi$) is singular in $L_\lambda$. But I guess it’s not clear how the singularity of $\kappa^{L_\lambda}$ in $L_\lambda$ might be related to singularity of $\kappa$ in $V$.

    Maybe the set-theorists on MO would have a thought.

15. • CommentRowNumber15.
    • CommentAuthorSridharRamesh
    • CommentTimeJan 11th 2011
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    Author: SridharRamesh
    Format: TextWell, of course, there are unhelpful, "cheating" ways of picking phi such as to achieve the undecidability; for example, defining phi(k) as "k is the lowest limit cardinal such that k is regular if and only if there are weakly innaccessible cardinals". So we ought to constrain the question a bit further, perhaps, though I'm not sure how exactly to formalize what counts as "cheating".

    Well, of course, there are unhelpful, "cheating" ways of picking phi such as to achieve the undecidability; for example, defining phi(k) as "k is the lowest limit cardinal such that k is regular if and only if there are weakly innaccessible cardinals". So we ought to constrain the question a bit further, perhaps, though I'm not sure how exactly to formalize what counts as "cheating".
16. • CommentRowNumber16.
    • CommentAuthorTobyBartels
    • CommentTimeJan 11th 2011
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    Author: TobyBartels
    Format: MarkdownItexI edited the offending remark in light of Sridhar's comment.

    I edited the offending remark in light of Sridhar’s comment.

17. • CommentRowNumber17.
    • CommentAuthorMike Shulman
    • CommentTimeJan 11th 2011
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    Author: Mike Shulman
    Format: MarkdownItexI guess one way to "prevent cheating" might be to require that the definition be absolute for some class of submodels, so that it "really defines" some particular cardinal. If we assume GCH, so that any limit cardinal is a strong limit and any weaky inaccessible is inaccessible, then I think any limit cardinal whose definition is absolute for $V_\lambda$ whenever $\lambda$ is inaccessible must be singular, by the argument I gave above. Namely, if $\lambda$ is the smallest inaccessible, then $\kappa$ is the same as $\kappa^{V_\lambda}$, which is singular in $V_\lambda$ since $V_\lambda$ contains no weak inaccessibles, and singularity in $V_\lambda$ should imply singularity in V.

    I guess one way to “prevent cheating” might be to require that the definition be absolute for some class of submodels, so that it “really defines” some particular cardinal. If we assume GCH, so that any limit cardinal is a strong limit and any weaky inaccessible is inaccessible, then I think any limit cardinal whose definition is absolute for $V_\lambda$ whenever $\lambda$ is inaccessible must be singular, by the argument I gave above. Namely, if $\lambda$ is the smallest inaccessible, then $\kappa$ is the same as $\kappa^{V_\lambda}$, which is singular in $V_\lambda$ since $V_\lambda$ contains no weak inaccessibles, and singularity in $V_\lambda$ should imply singularity in V.

18. • CommentRowNumber18.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 17th 2020
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    Author: Todd_Trimble
    Format: MarkdownItexAt [[small category]] it is written: $C$ is essentially $U$-small if there is a bijection from its set of morphisms to an element of $U$ (the same for the set of objects follows); this condition is non-evil. I don't understand how this is non-evil. Can't we just add in more than $U$ many isomorphic copies of an object to get an equivalent category?

    At small category it is written: $C$ is essentially $U$-small if there is a bijection from its set of morphisms to an element of $U$ (the same for the set of objects follows); this condition is non-evil.

    I don’t understand how this is non-evil. Can’t we just add in more than $U$ many isomorphic copies of an object to get an equivalent category?