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.

]]>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 $Set^{op}$ if and only if there exist *no* measurable cardinals. For any set $A$, the canonical colimit over this subcategory in $Set^{op}$ turns out to be essentially the set of countably additive measures on $A$, hence isomorphic to $A$ iff $A$ admits no countably additive measure. Then the extension to higher cardinals is fairly straightforward.

But it turns out that $Set^{op}$ 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…)

]]>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 $Set^{op}$ is the other sort of category, and I think there are mathematical consequences of that. For instance, would you expect $Set^{op}$ to have a small dense subcategory? If $Set$ is a category of spaces, so that $Set^{op}$ is a category of algebras, then I would see nothing wrong with its having a small dense subcategory; but if $Set$ is a category of algebras, so that $Set^{op}$ is a category of spaces, then I would expect it probably not to. But it turns out that $Set^{op}$ 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 $Set$ is a category of spaces or a category of algebras. (-:

Yes, that’s quite right: $FinSet$ *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. :-)

]]>Todd is helping me understand opposite categories beginning with $FinSet^{op}$ 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

spacesas 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 $FinSet$ is the category of *spaces* opposite to the category of finite Boolean algebras in the sense of space and quantity.