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    • CommentRowNumber1.
    • CommentAuthorEric
    • CommentTimeMay 8th 2010

    Todd is helping me understand opposite categories beginning with FinSet opFinSet^{op} here.

    This discussion helped prompt some improvement of the page opposite category. When I look at that page now, I see the statement:

    The idea of noncommutative geometry is essentially to define a category of spaces as the opposite category of a category of algebras.

    This reminded me of a remark I made in the “Forward” to a paper I wrote back in 2002, so I’ve now itexified that “Foreward” here:

    Noncommutative Geometry and Stochastic Calculus

    By the way, this also suggests that the category FinSetFinSet is the category of spaces opposite to the category of finite Boolean algebras in the sense of space and quantity.

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 8th 2010
    • (edited May 8th 2010)

    Yes, that’s quite right: FinSetFinSet is a category of spaces opposite to… But they are not the most interesting spaces in the world, being finite sets with the discrete topologies.

    Nevertheless, your comment is quite a propos. What we are doing at “opposite of finite sets” is a ’baby’ version of Stone duality, which posits a general duality between general Boolean algebras (not just finite ones) and so-called ’Stone spaces’. This was perhaps the first theorem of its kind, which went on to be extrapolated to Gelfand-Naimark duality (on maximal ideal spectra for C*-algebras), and thence to the more sophisticated noncommutative geometries.

    I had actually begun a series of posts (over on my all-but-defunct blog with Vishal Lama) on Stone duality… but maybe you shouldn’t read that right away, Eric, as we’re carrying on this quasi-Socratic (or Moore method) experiment. :-)

    • CommentRowNumber3.
    • CommentAuthorMike Shulman
    • CommentTimeMay 8th 2010

    It’s a very interesting question to me of whether the category of (perhaps finite) sets should be regarded as a category of spaces or as a category of algebras. On the one hand, you can think of a set as a space with a discrete topology. But on the other hand, you can think of it as an algebra with no operations.

    Of course whichever position one takes will suggest that Set opSet^{op} is the other sort of category, and I think there are mathematical consequences of that. For instance, would you expect Set opSet^{op} to have a small dense subcategory? If SetSet is a category of spaces, so that Set opSet^{op} is a category of algebras, then I would see nothing wrong with its having a small dense subcategory; but if SetSet is a category of algebras, so that Set opSet^{op} is a category of spaces, then I would expect it probably not to. But it turns out that Set opSet^{op} has a small dense subcategory if and only if there does not exist a proper class of measurable cardinals. So whether or not you believe in measurable cardinals could be seen as taking a stand on whether SetSet is a category of spaces or a category of algebras. (-:

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 8th 2010

    But it turns out that Set opSet^{op} has a small dense subcategory if and only if there does not exist a proper class of measurable cardinals.

    Huh! That’s very interesting. Where can I read about that? (If it’s in one of your papers on set theory and category theory, then I’ll be a little bit embarrassed that I didn’t pick up on that, but go ahead and say…)

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeMay 8th 2010

    Where can I read about that?

    It’s Theorem A.5 in Adamek&Rosicky, “Locally presentable and accessible categories.” I think it’s easier to understand if you first prove that the full subcategory on \mathbb{N} is dense in Set opSet^{op} if and only if there exist no measurable cardinals. For any set AA, the canonical colimit over this subcategory in Set opSet^{op} turns out to be essentially the set of countably additive measures on AA, hence isomorphic to AA iff AA admits no countably additive measure. Then the extension to higher cardinals is fairly straightforward.

    • CommentRowNumber6.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 8th 2010

    Thanks, Mike! It’s very strange to me that I haven’t looked into their appendix before, because it’s very well-written and a lot of stuff of interest to me. One thing I like about it is that it is written from a structuralist point of view, which I find immensely easier to follow than more materialist accounts.